J,   Henry  Senger 


Elements  of  Algebra. 


BY 

G.  A.  WENTWOKTH,  A.M., 

PROFESSOR  OF   MATHEMATICS   IN   PHILLIPS   EXETER  ACADEMY. 


BOSTON: 

PUBLISHED  BY  GINN  &  HEATH. 

1881. 


Entered  according  to  Act  of  Congress,  in  fhe  year  1881,  by 

G.  A.  Wentworth, 
in  the  office  of  the  Librarian  of  Congress,  at  Washington. 


IN  MEMORIAM 


GiNN  &  Heath: 

J.  S.  Gushing,  Printer,  i6  Hawley  Street, 

Boston. 


PREFACE, 


THE  single  aim  in  writing  thia  volume  has  been  to  make  an 
Algebra  which  the  beginner  would  read  with  increasing  in- 
terest, intelligence,  and  power.  The  fact  has  been  kept  constantly 
in  mind  that,  to  accomplish  this  object,  the  several  parts  must  be 
presented  so  distinctly  that  the  pupil  will  be  led  to  feel  that  he 
is  mastering  the  subject.  Originality  in  a  text-book  of  this  kind 
is  not  to  be  expected  or  desired,  and  any  claim  to  usefulness  must 
be  based  upon  the  method  of  treatment  and  upon  the  number 
and  character  of  the  examples.  About  four  thousand  examples 
have  been  selected,  arranged,  and  tested  in  the  recitation-room, 
and  any  found  too  difficult  have  been  excluded  from  the  book. 
The  idea  has  been  to  furnish  a  great  number  of  examples  for  prac- 
tice, but  to  exclude  complicated  problems  that  consume  time  and 
energy  to  little  or  no  purpose. 

In  expressing  the  definitions,  particular  regard  has  been  paid  to 
brevity  and  perspicuity.  The  rules  have  been  deduced  from  pro- 
cesses immediately  preceding,  and  have  been  written,  not  to  be 
committed  to  memory,  but  to  furnish  aids  to  the  student  in  fram- 
ing for  himself  intelligent  statements  of  his  methods.  Each  principle 
has  been  fully  illustrated,  and  a  sufficient  number  of  problems  has 
been  given  to  fix  it  firmly  in  the  pupil's  mind  before  he  proceeds 
to  another.  Many  examples  have  been  worked  out,  in  order  to 
exhibit  the  best  methods  of  dealing  with  different  classes  of  prob- 
lems and  the  best  arrangement  of  the  work  ;  and  such  aid  has 
been  given  in  the  statement  of  problems  as  experience  has  shown 

S2S477 


IV  PREFACE. 


to  be  necessary  for  the  attainment  of  the  best  results.  General 
demonstrations  have  been  avoided  whenever  a  particular  illustra- 
tion would  serve  the  purpose,  and  the  application  of  the  principle 
to  similar  cases  was  obvious.  The  reason  for  this  course  is,  that 
the  pupil  must  become  familiar  with  the  separate  steps  from  par- 
ticular examples,  before  he  is  able  to  follow  them  in  a  general 
demonstration,  and  to  understand  their  logical  connection. 

It  is  presumed  that  pupils  will  have  a  fair  acquaintance  with 
Arithmetic  before  beginning  the  study  of  Algebra;  and  that  suffi- 
cient time  will  be  afforded  to  learn  the  language  of  Algebra,  and 
to  settle  the  principles  on  which  the  ordinary  processes  of  Algebra 
are  conducted,  before  attacking  the  harder  parts  of  the  book.  "Make 
haste  slowly"  should  be  the  watchword  for  the  early  chapters. 

It  has  been  found  by  actual  trial  that  a  class  can  accomplish 
the  whole  work  of  this  Algebra  in  a  school  year,  with  one  reci- 
tation a  day ;  and  that  the  student  will  not  find  it  so  difficult  as 
to  discourage  him,  nor  yet  so  easy  as  to  deprive  him  of  the  re- 
wards of  patient  and  successful  labor.  At  least  one-third  of  the 
year  is  required  to  reach  the  chapter  on  Fractions ;  but,  if  the  first 
hundred  pages  are  thoroughly  mastered,  rapid  and  satisfactory 
progress  will  be  made  in  the  rest  of  the  book. 

Particular  attention  should  be  paid  to  the  chapter  on  Factoring; 
for  a  thorough  knowledge  of  this  subject  is  requisite  to  success  in 
common  algebraic  work. 

Attention  is  called  to  the  method  of  presenting  Choice  and  Chance. 
The  accomplished  mathematician  may  miss  the  elegance  of  the  gen- 
eral method  usually  adopted  in  Algebras ;  but  it  is  believed  that 
this  mode  of  treatment  will  furnish  to  the  average  student  the  only 
way  by  which  he  can  arrive  at  an  understanding  of  the  principles 
underlying  these  difficult  subjects.  In  the  preparation  of  these 
chapters  the  author  has  had  the  assistance  and  cooperation  of  G.  A. 
Hill,  A.M.,  of  Cambridge,  Mass.,  to  whom  he  gratefully  acknowl- 
edges his  obligation. 


PREFACE.  V 

The  materials  for  this  Algebra  have  been  obtained  from  English, 
German,  and  French  sources.  To  avoid  trespassing  upon  the 
works  of  recent  American  authors,  no  American  text-book  has 
been  consulted. 

The  author  returns  his  sincere  thanks  for  assistance  to  Rev.  Dr. 
Thomas  Hill ;  to  Professors  Samuel  Hart  of  Hartford,  Ct. ;  C.  H. 
Judson  of  Greenville,  S.C. ;  0.  S.  Westcott  of  Racine,  Wis. ;  G.  B. 
Halsted  of  Princeton,  N.J. ;  M.  W.  Humphreys  of  Nashville,  Tenn. ; 
W.  LeConte  Stevens  of  New  York,  N.Y.;  G.  W.  Bailey  of  New 
York,  N.Y. ;  Robert  A.  Benton  of  Concord,  N.H. ;  and  to  Dr.  D.  F. 
Wells  of  Exeter.  He  has  also  the  pleasure  of  expressing  his  obli- 
gations to  Messrs.  J.  S.  Gushing  and  F.  E.  Bartley,  to  whose  superior 
taste  and  judgment  the  typographical  excellence  of  this  book  is  due. 

There  will  be  two  editions  of  the  Algebra :  one  with  the  answers 
to  the  problems  boUnd  at  the  end  of  the  volume,  and  the  other 
without  answers.  Answers,  however,  b»und  separately,  may  be 
obtained.  The  edition  with  answers  bound  separately  is  strongly 
recommended. 

Any  corrections  or  suggestions  relating  to  the  work  will  be  thank- 
fully received. 

G.  A.  WENTWORTH. 
Phillips  Exeter  Academy, 
May,  1881. 


ELEMEISTTS    OF    ALGEBEA. 


CHAPTER;!,  '\\,AV  '-.'^ 
Quantity  and  Number. 

1.  "Whatever  may  be  regarded  as  being  made  up  of 
parts  like  the  whole  is  called  a  Quantity.  »  *• 

2.  To  measure  a  quantity  of  any  kind  is  to  find  how 
many  times  it  contains  another  hnown  quantity  of  the 
same  hind. 

3.  A  hnown  quantity  which  is  adopted  as  a  standard  for 
measuring  quantities  of  the  same  kind  is  called  a  Unit. 
Thus,  the  foot,  the  pound,  the  dollar,  the  day,  are  units 
for  measuring  distance,  weight,  money,  time. 

4.  A  Number  arises  from  the  repetitions  of  the  unit  of 
measure,  and  shows  how  Tnany  times  the  unit  is  contained 
in  the  quantity  measured. 

6.  When  a  quantity  is  measured,  the  result  obtained  is 
expressed  by  prefixing  to  the  name  of  the  unit  the  number 
which  shows  how  many  times  the  unit  is  contained  in  the 
quantity  measured ;  and  the  two  combined  denote  a  quan- 
tity expressed  in  units.  Thus,  7  feet,  8  pounds,  9  dollars, 
10  days,  are  quantities  expressed  in  their  respective  units. 


ALGEBRA. 


When  a  question  about  a  quantity  includes  tlie  unit,  the 

tyt^ep  answer  is  &.^7mmber;  when  it  does  not  include  the  unit, 

c^gl^    the  answer  is  a  quantity.     Thus,  if  a  man  who  has  fifteen 

bushels  of  wheat  be  asked  how  many  bushels  of  wheat  he 

has,  the  answer  is  the  number^  fifteen ;  if  he  be  asked  how 

rmtrih  wheat  he  Jiass,  the  answer  is  the  quantity,  fifteen 

bushels. 

'   ,      K  immber  answers  the  question,  How  many  ?  a  quantity, 

the  question,  Hov^  much  ? 

Numbers. 

6.  The  symbols  which  Aritlmietio  employs  to  represent 
numbers  are  the  figures  0,  1,'  2,  3,  4,  5,  6,  7,  8,  9.  The 
natural  series  of  numbers  begins  with  0 ;  each  succeeding 
number  is  obtained  by  adding  one  to  the  preceding  number, 
and  the  series  is  infinite. 

7.  Besides  figures,  the  chief  symbols  used  in  Arithmetic 
are : 

+  (read,  plus),  the  sign  of  addition. 

—  (read,  minus),  the  sign  of  subtraction. 

X  (read,  multiplied  into),  the  sign  of  multiplication. 

-V-  (read,  divided  by),  the  sign  of  division. 

=  (read,  is  equal  to),  the  sign  of  equality. 

Exercise.  —  Read : 

7  +  12  =  19.  8  +  3-5=    20-15  +  1. 

9-   4=    5.  24  +  6=    10  X    3. 

6x    4  =  24.  14-7  +  5=     6x2. 

48 -V-   3  =  16.  9x5  =  180--   4. 

8.  Any  figure,  or  combination  of  figures,  as  7,  28,  346, 
has  one,  and  only  one,  value.     That  is,  figures  represent 


NUMBERS. 


^particular  numbers.      But  numbers  possess  many  general 
^^^QpertieSy  which  are  true,  not  only  of  a  particular  number, 
but  of  all  numbers. 

Thus,  the  sum  of  12  and  8  is  20,  and  the  difference  be- 
tween 12  and  8  is  4.  Their  sum  added  to  their  difference 
is  24,  which  is  twice  the  greater  number.  Their  differ- 
ence taken  from  their  sum  is  16,  which  is  twice  the  smaller 
number. 

9.  As  this  is  true  of  any  two  numbers,  we  have  this  gen- 
eral property :    The  sum  of  two  numbers  added  to  their  differ- 
ence is  twice  the  greater  number ;  the  difference  of  two  numbers        I 
taken  from  their  sum  is  twice  the  smaller  number.     Or,  -* 

1.  (greater  number  +  smaller  number)  +  (greater  number 

—  smaller  number)  =  twice  greater  number. 

2.  (greater  number  +  smaller  number)  —  (greater  number 

—  smaller  number)  =  twice  smaller  number. 

But  these  statements  may  be  very  much  shortened ;  forN 
as  gre^iter  number  and  smaller  number  may  mean  any  two 
numbers,  two  letters,  as  a  and  b,  may  be  used  to  represent 
them;  and  2a  may  represent  twice  the  greater,  and  2b 
twice  the  smaller.     Then  these  statements  become : 

1.  {a  +  b)-^(a-b)  =  2a. 

2.  {a-\-b)-(a-b)  =  2b. 

In  studying  the  general  properties  of  numbers,  letters 
may  represent  any  numerical  values  consistent  with  the 
conditions  of  the  problem. 

10.  It  is  also  convenient  to  use  letters  to  denote  numbers 
which  are  unknown,  and  which  are  to  be  found  from  certain 
given  relations  to  other  known  numbers. 


ALGEBRA. 


Thus,  the  solution  of  the  problem,  "  Find  two  numbers 
such  that,  when  the  greater  is  divided  by  the  less,  the  quo- 
tient is  4,  and  the  remainder  3 ;  and  when  the  sum  of  the 
two  numbers  is  increased  by  38,  and  the  result  divided  by 
the  greater  of  the  two  numbers,  the  quotient  is  2  and  the 
remainder  2,"  is  much  simplified  by  the  use  of  letters  to 
represent  the  unknown  numbers. 

11.  The  science  which  employs  letters  in  reasoning  about 
numbers,  either  to  discover  their  general  properties,  or  to 
find  the  value  of  an  imJmoum  number^  from  its  relations 
to  known  numbers,  is  called  Algebra. 

Algebeaic  Numbers. 

12.  There  are  quantities  which  stand  to  each  other  in 
such  opposite  relations  that,  when  we  combine  them,  they 
cancel  each  other  entirely  or  in  part.  Thus,  six  dollars 
gain  and  six  dollars  loss  just  cancel  each  other ;  but  ten 
dollars  gain  and  six  dollars  loss  only  cancel  each  other  in 
part.  For  the  six  dollars  loss  will  cancel  six  dollars  of  the 
gain  and  will  leave  four  dollars. 

An  opposition  of  this  kind  exists  in  assets  and  debts,  in 
income  and  outlay,  in  motion  forwards  and  backwards,  in 
motion  to  the  right  and  to  the  left,  in  time  before  and  after 
a  fixed  date,  in  the  degrees  above  and  below  zero  on  a 
thermometer. 

From  this  relation  of  quantities  a  question  often  arises 
which  is  not  considered  in  Arithmetic ;  namely,  the  sub- 
tracting of  a  greater  number  from  a  smaller.  This  cannot 
be  done  in  Arithmetic,  for  the  real  nature  of  subtraction 
consists  in  counting  backwards,  along  the  natural  series  of 
numbers.  If  we  wish  to  subtract  four  from  six,  we  start 
at  six  in  the  natural  series,  count  four  units  backwards,  and 


ALGEBRAIC    NUMBERS. 


arrive  at  two,  the  difference  sought.  If  we  subtract  six 
from  six,  we  start  at  six  in  the  natural  series,  count  six 
units  backwards,  and  arrive  at  zero.  If  we  try  to  subtract 
nine  from  six,  we  cannot  do  it,  because,  when  we  have 
counted  backwards  as  far  as  zero,  the  natural  series  of 
nuTTibers  comes  to  an  end. 

I 

13.  In  order  to  subtract  a  greater  mimber  from  a  smaller 
it  is  necessary  to  assume  a  new  series  of  numbers,  beginning 
at  zero  and  extending  to  the  left  of  zero.  The  series  to  the 
left  of  zero  must  ascend  from  zero  by  the  repetitions  of  the 
unit,  precisely  like  the  natural  series  to  the  right  of  zero ; 
and  the  opposition  between  the  right-hand  series  and  the 
left-hand  series  must  be  clearly  marked.  This  opposition 
is  indicated  by  calling  every  number  in  the  right-hand 
series  a  positive  number,  and  prefixing  to  it,  when  written, 
the  sign  +  ;  and  by  calling  every  number  in  the  left-hand 
series  a  negative  number,  and  prefixing  to  it  the  sign  — . 
The  two  series  of  numbers  will  be  written  thus : 


-4,-3,-2,-1,  0,-M,+2,+3,-f4, 


If,  now,  we  wish  to  subtract  9  from  6,  we  begin  at  6  in 
the  positive  series,  count  nine  units  in  the  negative  direction 
(to  the  left),  and  arrive  at  —  3  in  the  negative  series.  That 
is,  6-9  =  -3. 

The  result  obtained  by  subtracting  a  greater  number 
from  a  less,  when  both  are  positive,  is  always  a  negative 
number. 

If  a  and  h  represent  any  two  numbers  of  the  positive 
series,  the  expression  a  —  h  will  denote  a  positive  number 
when  a  is  greater  than  h ;  will  be  equal  to  zero  when  a  is 
equal  to  h ;  will  denote  a  negative  number  when  a  is  less 
than  h. 

If  we  wish  to  add  9  to  —6^  we  begin  at  —6,  in  the 


(j  ALGEBRA. 

negative  series,  count  nine  units  in  the  positive  direction 
(to  the  right),  and  arrive  at  +  3,  in  the  positive  series. 

We  may  illustrate  the  use  of  positive  and  negative  num- 
bers as  follows : 

^^4 ^ -^0 

DA  a 

Suppose  a  person  starting  at  A  walks  20  feet  to  the  right 
of  A,  and  then  returns  12  feet,  where  will  he  be  ?  Answer  : 
at  C,  a  point  8  feet  to  the  right  of  A.  That  is,  20  feet  — 
12  feet  =  8  feet ;  or,  20  -  12  =  8. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  20  feet,  where  will  he  be  ?  Answer :  at  A,  the 
point  from  which  he  started.     That  is,  20  —  20  =  0. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet,  and 
then  returns  25  feet,  where  will  he  now  be  ?  Answer :  at 
D,  a  point  5  feet  to  the  left  of  A.  That  is,  if  we  consider 
distance  measured  in  feet  to  the  left  of  A  as  forming  a 
negative  series  of  numbers,  beginning  at  A,  20  —  25  =  —  5. 
Hence,  the  phrase,  5  feet  to  the  left  of  A,  is  now  expressed 
by  the  negative  number  —  5. 

14.  Numbers  provided  with  the  sign  +  or  —  are  called 
algebraic  nuinbers.  They  are  unknown  in  Arithmetic,  but 
play  a  very  important^  part  in  Algebra.  In  contradistinc- 
tion, numbers  not  affecteji  by  the  signs  +  or  —  are  termed 
absolute  nniiibersr'" 

15.  Eyery  algebraic  number,  as  +4  or  —  4,  consists  of 
a  sign  +  or  —  and  the  absolute  value  of  the  number ;  in 
this  case  4.  The  sign  shows  whether  the  number  beloijgs 
to  the  positive  or  negative  series  of  numbers  ;  the  absolute 
value  shows  what  place  the  number  has  in  the  positive  or 
negative  series. 


FACTORS    AND    POWERS. 


16.  "When  no  sign  stands  before  a  number,  the  sign  +  is 
always  understood  ;  thus,  4  means  the  same  as  +4,  a  mieans 

'^  the  slime  as  -j-  a.     But  the  sign  —  is  neva'  omitted, 

17.  Two  numbers  which  have,  one  the  sign  +  and  the 
other  the  sign  — ,  are  said  to  have  U]dike_si^s.^ 

18.  Two  numbers  which  have  the  same^absolute  values, 
but  unlike  signs,  always  cancel  each  other  when  combined  ; 
thiis,  +  4  —  4  =  0 ,  +  a  -  a  =  0. 

19.  The  use  of  the  signs  +  and  — ,  to  indicate  addition 
and  subtraction,  must  be  carefully  distinguished  from  their 
use  to  indicate  in  which  series,  the  positive  or  the  negative, 
a  given  number  belongs.  In  the  first  sense,  they  are  signs 
of  operatixms,  and  are  common  to  both  Arithmetic  and  Al- 
gebra. In  the  second  sense,  they  are  signs  of  opposition^ 
and  are  employed  in  Algebra  alone. 

Factors  and  Powers. 

20.  When  a  number  consists  of  the  product  of  two  or 
more  numbers,  each  of  these  numbers  is  called  a  factor  of 
the  product. 

When  these  numbers  are  denoted  by  letters,  the  sign  X 
is  omitted ;  thus,  instead  of  a  X  5,  we  write  ab ;  instead 
of  a  X  5  X  c,  we  write  ahc. 

The  expression  abc  must  not  be  confounded  with  a+5  +  c; 
the  first  is  a  product,  the  second  is  a  sum.  If  a  =  2,  6  =  3, 
0  =  4:,  then 

a5c  =  2x  3x4  =  24; 

a  +  6  +  c  =  2  +  3  +  4=    9. 

21.  Factors  expressed  by  letters  are  called  literal  factors ; 
factors  expressed  by  figures  are  called  numerical  factors. 


8  ALGEBRA. 


22.  A  known  factor  of  a  product  whicli  is  prefixed  to  an- 
other factor,  to  show  how  many  times  that  factor  is  taken, 
is  called  a  coefficient.  Thus,  in  7c,  7  is  the  coefficient  of  c ; 
in  ^ax,  7  is  the  coefficient  of  ax,  or,  if  a  be  known,  7a  is 
the  coefficient  of  x.  When  no  numerical  coefficient  occurs 
in  a  product,  1  is  always  understood.  Thus,  ax  means  the 
same  as  \ax. 

23.  A  product  consisting  of  two  or  more  equal  factors  is 
called  a  power  of  that  factor. 

24.  The  index  or  exponent  of  a  power  is  a  small  figure  or 
letter  placed  at  the  right  of  a  number,  to  show  how  many 
times  the  number  is  taken  as  a  factor.  Thus,  a^,  or  simply 
a,  denotes  that  a  is  taken  once  as  a  factor  ;  d?  denotes  that 
a  is  taken  twice  as  a  factor  ;  a?  denotes  that  a  is  taken  three 
times  as  a  factor  ;  and  a**  denotes  that  a  is  taken  n  times  as 
a  factor.  These  are  read :  the  first  power  of  a  ;  the  second 
power  of  a  ;  the  third  power  of  a  ;  the  Tith  power  of  a. 

c?  is  written  instead  of  aaa. 

a"  is  written  instead  of  aaa,  etc.,  repeated  n  times. 
The  meaning  of  coefficient  and  exponent  must  be  care- 
fully distinguished.     Thus, 

4a  =a  +  a  +  «4-«; 
a^  =  aXaXaXa. 
Ifa==3,  4a  =3  +  3  +  3  +  3  =  12. 


25.  The  second  power  of  a  number  is  generally  called 
the  square  of  that  number ;  thus,  a^  is  called  the  square 
of  a,  because  if^a  denote  the  number  of  units  of  length  in 
the  side  of  a  square,  a^  denotes  the  number  of  units  of 
surface  in  the  square. 


ALGEBRAIC    SYMBOLS. 


The  third  power  of  a  number  is  generally  called  the  cube 
of  that  number ;  thus,  o?  is  called  the  cube  of  a,  because 
if  a  denote  the  number  of  units  of  length  in  the  edge  of  a 
cube,  a^  denotes  the  number  of  units  of  volume  in  the  cube. 


Algebeaic  Symbols. 

26.  Known  numbers  in  Algebra  are  denoted  by  figures 
and  by  the  first  letters  of  some  alphabet ;  as,  a,  b,  c,  etc.  ; 
d,  b\  c\  read  a  prime,  b  prime,  c  priyne,  etc. ;  «!,  ^j,  Ci,  read 
a  one,  b  one,  c  one. 

Unknown  numbers  are  generally  denoted  by  the  last  let- 
ters of  some  alphabet ;  as,  x,  y,  z,  x' ,  'i/ ,  /,  etc. 

27.  The  symbols  of  operations  are  the  same  in  Algebra 
as  in  Arithmetic.  One  point  of  difference,  however,  must 
be  carefully  observed.  When  a  symbol  of  operation  is  omit- 
ted in  the  notation  of  Arithmetic,  it  is  always  the  symbol 
of  addition ;  but  when  a  symbol  of  operation  is  omitted  in 
the  notation  of  Algebra,  it  is  always  the  symbol  of  mul- 
tiplication.    Thus,  456  means  400  +  50  +  6,  but  4  ab  means 

AxaXb;  4f  means  4  +  f ,  but  4^  means  4  X  ^• 

o  b 

28.  The  symbols  of  relation  are  =,  >,  <,  which  stand 

for  the  words,  "  is  equal  to,"  "  is  greater  than,"  and  "  is  less 
than,"  respectively. 

29.  The  symbols  of  aggregation  are  the  bar,  | ;  the  vin- 
culum,   ;   the  parenthesis,  (  )  ;  the  bracket,  [  ] ;   and 

the  brace,  {  } .     Thus,  each  of  the  expressions,    ,      ,  x-\~y, 

(a;  +  y),  [a;  +  y],  {x-^-y],  signifies  that  a; +  y  is  to  be  treated 
as  a  single  number. 


10  ALGEBRA. 


30.  The  symbols  of  oontiimatioii  are  dots,  ,  or  dashes, 

,  and  are  read,  "and  so  on." 

31.  The  symbol  of  deduction  is  .'.,  and  is  read,  "hence," 
or  "therefore." 

Algebraic  Expressions. 

32.  An  algebraic  expression  is  any  number  written  in 
algebraic  symbols.  Thus,  8  c  is  the  algebraic  expression 
for  8  times  the  number  denoted  by  c. 

lo?  —  2)ah  is  the  algebraic  expression  for  7  times  the 
square  of  the  number  denoted  by  a,  diminished  by  3  times 
the  product  of  the  numbers  denoted  by  a  and  h. 

33.  A  term  is  an  algebraic  expression  the  parts  of  which 
are  not  separated  by  the  signs  of  addition  or  subtraction. 
Thus,  3  «5  H-  5  :ry  is  a  term. 

34.  A  monomial  or  simple  expression  is  an  expression 
which  contains  only  one  term. 

35.  A  polynomial  or  compound  expression  is  an  expression 
which  contains  two  or  more  terms.  A  binomial  is  a  poly- 
nomial of  two  terms.  A  trinomial  is  a  polynomial  of  three 
terms. 

36.  Like  terms  are  terms  which  have  the  same  letters, 
and  the  corresponding  letters  affected  by  the  same  expo- 
nents. Thus,  la^ca^  and  —ba?co(^  are  like  ^ terms;  but 
la^cx^  and  —  bac^a?  are  unlike  terms. 

37.  The  dimensions  of  a  term  are  its  literal  factors. 

38.  The  degree  of  a  term  is  equal  to  the  number  of  its 
dimensions,  and  is  found  by  taking  the  sum  of  the  expo- 
nents of  its  literal  factors.  Thus,  ?>xy  is  of  the  second 
degree,  and  bx^y^  is  of  the  sixth  degree. 


AXIOMS.  11 


39.  A  polynomial  is  said  to  be  homogeneous  when  all  its 
terms  are  of  the  same  degree.  Thus,  l:i?—bji^y-\-xyz  is 
homogeneous,  for  each  term  is  of  the  third  degree. 

40.  A  polynomial  is  said  to  be  arranged  according  to  the 
powers  of  some  letter  when  the  exponents  of  that  letter 
either  descend  or  ascend  in  regular  order.  Thus,  Zaoi? 
—  4 hx^  —  Qax-{-Sb  is  arranged  according  to  the  descend- 
ing powers  of  x,  and  8  Z>  —  6  aa;  —  4  hx^  +  3  aa?  is  arranged 
according  to  the  ascending  powers  of  x. 

41.  The  numerical  value  of  an  algebraic  expression  is  the 
number  obtained  by  giving  a  particular  value  to  each  letter, 
and  then  performing  the  operations  indicated. 

42.  Two  numbers  are  reciprocals  of  each  other  when  their 

product  is  equal  to  unity.     Thus,  a  and  -  are  reciprocals. 

a 

Axioms. 

43.  1.  Things  which  are  equal  to  the  same  thing  are 
equal  to  each  other. 

2.  If  equal  numbers  be  added  to  equal  numbers,  the 
sums  will  be  equal. 

3.  If  equal  numbers  be  subtracted  from  equal  numbers, 
the  remainders  will  be  equal. 

4.  If  equal  numbers  be  multiplied  into  equal  numbers, 
the  products  will  be  equal. 

5.  If  equal  numbers  be  divided  by  equal  numbers,  the 
quotients  will  be  equal. 

6.  If  the  same  number  be  both  added  to  and  subtracted 
from  another,  the  value  of  the  latter  will  not  be  altered. 

7.  If  a  number  be  both  multiplied  and  divided  by  an- 
other, the  value  of  the  former  will  not  be  altered. 


12  ALGEBRA. 


Exercise  I. 

If  a  =  l,  b  =  2,  c  =  S,  d  =  4:,  e  =  5, /=  0,  find  the  nu- 
merical values  of  the  following  expressions  : 

-.     n      loz-io        or  .    4:ac  ,  8bc     bed 

1.  9  a  +  2  6  +  3  (?  —  2/.  4.   — — 1 ; • . 

ode 

2.  4e~3a-35  +  5^.  5.   1e  +  bcd-^^. 

2ae 

3.  Sabc  —  bcd-i-9cde  —  def.     6.    abc^ -\-bcd'^  —  dec? -\-p, 

8a^  +  35^  ■  4g^  +  6Z>^      g^  +  6^^ 
9.^.  11.        ^^+^" 


Z*"'  b^  +  d^-b(£ 

10.  ^4±i:      12.    ^^-^^ 


-b'  e'  +  ed+d^ 

When  a  term  contains  the  signs  X  or  ~,  the  operation 
indicated  by  these  signs  must  be  performed  before  the 
operations  of  addition  and  subtraction. 

Simplify  the  following  expressions  : 

13.  100  +  80--4.  15.   25  +  5X4-10-V-5. 

14.  75-25x2.  16.   24-5x4-hl0  +  3. 

17.   (24-5)x(4--10  +  3). 

Find  the  numerical  value  of  the  following  expressions,  in 
which  a  =  2,  5  =  10,  a;  =  3,  y  =  5  : 

18.  ;ry  +  4ax2.  20.   3a;  +  7y-v- 7  +  a  X  y. 

19.  xy  —  lbb-^b.  21.    ^b —  Sy  ^2y  Xb  —  2b. 


ALGEBRAIC   NOTATION.  13 

22.  {<oh~^y)^2yXh  +  2h. 

23.  (6  5-8y)--(2yX^')  +  25. 

24.  6b-(8y^2y)xb-2b. 

25.  6b~(b-y)-Sx-{-bxi/-^10a.         )^^ 

Algebraic  Notation. 

26.  Express  the  sum  of  a  and  b. 

27.  Express  the  double  of  :r.        -'>'>vN 

28.  By  how  much  is  a  greater  than  5  ?  ^  "^^ . 

29.  If  a;  be  a  whole  number,  what  is  the  next  number  above 

it? 

30.  Write  five  numbers  in  order  of  magnitude,  so  that  x 

shall  be  the  middle  number.       .  -y;  -^  j .    >f    / 

31.  What  is  the  sum  of  x_  +  x-{-  x  -f written  a  times  ? 

32.  If  the  product  be  xy  and  the  multiplier  x,  what  is  the 

multiplicand  ? 

33.  A  man  who  has  a  dollars  spends  b  dollars ;  how  many- 

dollars  has  he  left  ?  i       . 

34.  A  regiment  of  men  can  be  drawn  up  in  a  ranks  of  b  men 

each,  and  there  are  c  men  over ;  of  how  many  men 


does  the  regiment  consist  ? 


35.  Write,  the  sum  of  x  and  y  divided  by  c  is  equal  to  the 

product  of  a,  b,  and  ??i,  diminished  by  six  times  c,  and 
increased  by  the  quotient  of  a  divided  by  the  sum  of 
X  and  y. 

36.  Write,  six  times  the  square  of  w,  divided  by  m  minus  a, 

increased  by  five  b  into  the  expression  c  plus  d 
minus  a. 

37.  Write,  four  times  the  fourth  power  of  a,  diminished  by 

five  times  the  square  of  a  into  the  square  of  b,  and 
increased  by  three  times  the  fourth  power  of  b. 


14  ALGEBEA. 


Exercise  II. 

That  the  beginner  may  see  how  Algebra  is  employed  in 
the  solution  of  problems,  the  following  simple  exercises  are 
introduced : 

1.  John  and  James  together  have  $6.     James  has  twice 

as  much  as  John.     How  much  has  each  ? 

Let  X  denote  the  number  of  dollars  John  has. 

Then  2  a;  =  number  of  dollars  James  has, 

and  x-\-2x  =  number  of  dollars  both  have. 

But  6  =  number  of  dollars  both  have  ; 

-  .-.  x-\-2x=Q, 
or  3rB  =  6. 

and  x  =  2. 

Therefore,  John  has  $2,  and  James  has  $4. 

2.  A  stick  of  timber  40  feet  long  is  sawed  in  two,  so  that 

one  part  is  two-thirds  as  long  as  the  other.    Kequired 

the  length  of  each  part. 

Let  3  x  denote  the  number  of  feet  in  the  longer  part. 

Then  2x  =  number  of  feet  in  the  shorter  part, 

and        3  a;  +  2  a;  =  number  of  feet  in  both  together. 

But  40  =  number  of  feet  in  both  together ; 

.-.  3a;+2a;=40, 
or  5a;  =  40, 

and  x=    8. 

Therefore,  the  longer  part,  or  ^x,  is  24  feet  long  ;  and' 
the  shorter,  or  2x,  is  16  feet. 

Note.  The  unit  of  the  quantity  sought  is  always  given,  and  only 
the  nninter  of  such  units  is  required.  Therefore,  x  must  never  be  put 
for  money,  length,  time,  weight,  etc.,  but  always  for  the  required 
number  of  specified  units  of  money,  length,  time,  weight,  etc. 

The  beginner  should  give  particular  attention  to  this^  caution. 


PROBLEMS.  15 

3.  The  greater  of  two  numbers  is  six  times  the  smaller, 

and  their  sum  is  35.     Required  the  numbers. 

4.  Thomas  had  75  cents.     After  spending  a  part  of  his 

money,  he  found  he  had  twice  as  much  left  as  he 
had  spent.     How  much  had  he  spent  ? 

5.  A  tree  75  feet  high  was  broken,  so  that  the  part  broken 

off  was  four  times  the  length  of  the  part  left  standing. 
Required  the  length  of  each  part. 

6.  Four  times  the  smaller  of  two  numbers  is  three  times 

the  greater,  and  their  sum  is  G3.  Required  the  num- 
bers. 

7.  A  farmer  sold  a  sheep,  a  cow,  and  a  horse,  for  $216. 

He  sold  the  cow  for  seven  times  as  much  as  the 
sheep,  and  the  horse  for  four  times  as  much  as  the 
cow.     How  much  did  he  get  for  each  ? 

8.  George  bought  some  apples,  pears,  and  oranges,  for  91 

cents.  He  paid  twice  as  much  for  the  pears  as  for 
the  apples,  and  twice  as  much  for  the  oranges  as  for 
the  pears.    How  much  money  did  he  spend  for  each  ? 

9.  A  man  ];)ought  a  horse,  wagon,  and  harness,  for  $  350. 

He  paid  for  the  horse  four  times  as  much  as  for  the 
harness,  and  for  the  wagon  one-half  as  much  as  for 
the  horse.     What  did  he  pay  for  each  ? 

10.  Distribute  $3  among  Thomas,  Richard,  and  Henry,  so 

that  Thomas  and  Richard  shall  each  have  twice  as 
much  as  Henry. 

11.  Three  men,  A,  B,  and  C,  pay  $1000  taxes.     B  pays  4 

times  as  much  as  A,  and  C  an  amount  equal  to  the 
sum  of  what  the  other  two  pay.  How  much  does 
each  pay  ? 


CHAPTER  II. 

Addition  and  Subtraction. 

44.  An  algebraic  number  which  is  to  be  added  or  sub- 
tracted is  often  inclosed  in  a  parenthesis,  in  order  that  the 
signs  +  and  —  which  are  used  to  distinguish  positive  and 
negative  numbers  may  not  be  confounded  with  the  +  and 
—  signs  that  denote  the  operations  of  addition  and  subtrac- 
tion. Thus,  +  4  +  (—  3)  expresses  the  sum,  and  -f  4  —  (—  3) 
expresses  the  difference,  of  the  numbers  +  4  and  —  3. 

45.  In  order  to  add  two  algebraic  numbers,  we  begin  at 
the  place  in  the  series  which  the  first  number  occupies,  and 
count,  in  the  direction  indicated  by  the  sign  of  the  second 
number,  as  many  units  as  are  equal  to  the  absolute  value 
of  the  second  number.  Thus,  the  sum  of  +  4  +  (+  3)  is 
found  by  counting  from  +4  three  units  in  the  positive 
direction,  and  is,  therefore,  +  7 ;  the  sum  of  +  4  +  (—  3)  is 
found  by  counting  from  +4  three  units  in  the  negative 
direction,  and  is,  therefore,  +1. 

In  like  manner,  the  sum  of  —  4  +  (+  3)  is  —  1,  and  the 
sum  of  -  4  +  (-  3)  is  -  7.     That  is, 

(1)  +4  +  (+3)  =  7;  (3)   -4  +  (+3)  =  -l; 

(2)  +4  +  (-3)  =  l;  (4)   -4  +  (-3)  =  -7. 

I.  Therefore,  to  add  two  numbers  with  like  signs,  find 
the  sum  of  their  absolute  values,  and  prefix  the  com^mon  sign 
to  the  sum. 

II.  To  add  two  numbers  with  unlike  signs,  find  the  differ- 
ence of  their  absolute  values,  and  prefix  the  sign  of  the  greater 
number  to  the  difference. 


ADDITION.  17 


Exercise  III. 

1.  +16  +  (-ll)=  3.  +68  +  (-79)  = 

2.  -15  +  (-25)=  4.  -7  +  (+4)  = 

5.  +33  +  (+18)  = 

6.  +378 +  (+709) +  (-592)  = 

7.  A  man  has  $  5242  and  owes  $2758.     How  much  ia  he 

worth  ? 

8.  The  First  Punic  War  began  B.C.  264,  and  lasted  23 

years.     When  did  it  end? 

9.  Augustus  Caesar  was  born  B.C.  63,  and  lived  77  years. 

When  did  he  die  ? 
10.   A  man  goes  65  steps  forwards,  then  37  steps  backwards, 
then  again  48  steps  forwards.     How  many  steps  did 
he  take  in  all  ?     How  many  steps  is  he  from  where 
he  started  ? 

Addition  of  Monomials.  . 

46.  If  a  and  h  denote  the  absolute  values  of  any  two 
numbers,  1,  2,  3,  4  (§  45)  become :  . 

(1)  +a  +  (+Z»)  =  a  +  5;  (3)    -a  +  (+^»)=_a  +  5; 

(2)  -\-cu  +  {-h)  =  a-h\  (4)    -a-\-{-h)=-a-h. 

Therefore,  to  add  two  terms,  write  them  one  after  the  other 
with  unchanged  signs. 

It  should  be  noticed  that  the  order  of  the  terms  is  im- 
material. Thus,  -\-a  —  h  =  —  h-\-a.  Ifa  =  8  and  b  =  12, 
the  result  in  either  case  is  —  4. 

47.  3a  +  5a  +  2a  +  6a  +  a  =  17a. 
—  2c  —  ^c  —  c  —  4:C-Sc  =  —  18c. 

Therefore,  to  add  several  like  terms  which  have  the  same 


18  ALGEBRA. 


sign,  add  the  coefficients,  prefix  the  coTnmon  sign,  and  annex 

the  common  symbols, 

/ 

48.  7a-6a+lla  +  a-5a-2a  =  19a-13a  =  6a. 

—  3a  — 15a -7a  +  14a  — 2a  =  14a  — 27a  =  — 13a. 

Therefore,  to  add  several  like  terms  which  have  not  all 
the  same  sign,  find  the  difference  between  the  sum,  of  the 
positive  coefficients  and  the  sum  of  the  negative  coefficients, 
prefix  the  sign  of  the  greater  sum,  and  annex  the  common 
symbols. 

49.  5a-25  +  3a  =  8a-25. 

—  Zax-\-^y-\-^ax  —  4:C^^ax-\-Sy  —  4:C. 

Therefore,  to  add  terms  which  are  not  all  like  terms, 
combine  the  like  terms,  and  write  down  the  other  terms,  each 
preceded  by  its  proper  sign. 

Exercise  IV. 
L   5a&  +  (— 5aJ)=  Q  6.    7a5  +  (— 5a5)  = 

2.  Smx  +  (—  2mx)  =  7.    120 wy  +  (—  95my)  = 

3.  -lSmng-i-(-7mng)=     8.    -  S3  ab^ +(  11  ab'^) -= 

4.  -5x'  +  (+8a^)=  9.   -  75  rry  +  (+ 20  ^y)  = 

5.  25my'+(-18my^=      10.   +15aV  +  (-aV)  = 

11.  -b^m^  +  (+1b^m')  = 

12.  5a  +  (-3^)  +  (+4a)  +  (-7Z')  = 

13.  4:a^c+(~lOxyz')  +  (+e>a^c)  +  (-9xyz) 

+  (-lla'c)  +  (+20x2jz)^ 

14.  3^y  +  (-4aZ>)  +  (-2mn)  +  (+5:r2y) 


ADDITION.  19 


Addition  of  Polynomials. 

50.  Two  or  more  polynoniials  are  added  by  adding  their 
separate  terms. 

It  is  convenient  to  arrange  the  terms  in  columns,  so  that 
like  terms  shall  stand  in  the  same  column.     Thus, 

(I)2a^-3a25  +  4a52+    y    (2) -2x'y  -\-Qf-l 

-3a«+    an-2>ab^-U^  Qa^y  +2 

2a^-^-2a?h-\-^ah^-2,h^  a^y  -    f 

2a'  +  ^a'h  ^^«  -^^3/ zl 

-    a?y  +  2xf-{-b,/-\-l 

.  , ,  Exercise  V. 

Add: 

1.  5a  +  35  +  c,     3a  +  3J4-3^,     a4-32>  +  5c. 

2.  7a-45  +  c,     6a  +  36--5c,     -12a  +  4c?. 

3.  a-{-h  —  c,     h-\-c  —  a,     c-\-a  —  h,     a-\-b-~c. 

4.  a +  25  + 3c,     2a  —  h  —  2c,     h  —  a  —  c,     c  —  a  —  h. 

5.  a  — 25  +  3c  — 4c?,     3Z> -4c  + 5(^— 2a, 

5(?-6(^+3a-45,     7c?-4a  +  55 -4c?. 

6.  a;3_4a;2_|_5^_3^     2a;3_  7^_  7^_  14^_^  5^ 

7.  a;*-2a;3  +  3a,'2,     aj3  +  a:2_{_^^     4a;*  +  5:r3, 

2x'  +  ^x-4:,     -2>c(^-2x-b. 

8.  a«  +  3a62-3a25-^»^     2^3+ Sa^S  -  6a^.2_  7  j2^ 

a^-a¥  +  2h\ 

9.  2a5-3aar^  +  2a2;r,     12a^» -6a2^+ lOa^r^^ 

as^  —  Sab  —  ba^x. 


20  ALGEBRA. 


10.  c'-S(^  +  2e'-4:C+1,     2c^-i-3c^  +  2(^  +  5c-{-6, 

11.  Sci^  —  x^  +  xz  —  Sy^-{-4:7/z~z^,    —^x^  —  xy  —  xz-\-hyz, 

6:^-6y-62,     ^yz-byz-\-^z^, 
-^x'-\-f-\-%yz^Zz\ 

12.  7n^  —  ^m^n  —  ^m^v?,     -{-m^n^-{-m^n^— bm*n, 

2mn^  +  2n^-\-Sm\     —n^  +  2m^-{-1m'^n. 

Subtraction. 

51.  In  order  to  find  the  difference  between  two  algebraic 
numbers,  we  begin  at  the  place  in  the  series  ivhich  the  minu- 
end occupies,  and  count  in  the  direction  opposite  to  that  indi- 
cated by  the  sign  of  the  subtrahend  as  many  units  as  are 
equal  to  the  absolute  value  of  the  subtrahend. 

Thus,  the  difference  between  +  4  and  +  3  is  found  by 
counting  from  +  4  three  units  in  the  negative  direction,  and 
is,  therefore,  + 1 ;  the  difference  between  +  4  and  —  3  is 
found  by  counting  from  +  4  three  units  in  the  positive  direc- 
tion, and  is,  therefore,  +  7. 

In  like  manner,  the  difference  between  —  4  and  +  3  is 
—  7  ;  the  difference  between  —  4  and  —  3  is  —  1. 

Compare  these  results  with  results  obtained  in  addition : 


+  4  +  (-3)  =  l. 
+  4  +  (+3)-7. 
-4  +  (-3)  =  -7. 
-4  +  (+3)  =  -l. 


(1)  +4-(+3)  =  l 

(2)  +4-(-3)=:7 

(3)  _4-(+3)=-7 

(4)  _4-(-3)=:-l 

Or,     (1)  +4-(+3)  =  +  4  +  (-3). 

(2)  +4-(-3)-  +  4  +  (+3). 

(3)  -4-(+3)  =  -4  +  (-3). 

(4)  _4-(-3)=-4  +  (+3). 


SUBTRACTION.  21 


52.  From  (1)  and  (3),  it  is  evident  that  svhtracting  a 
positive  number  is  equivalent  to  adding  an  equal  negative 
number. 

From  (2)  and  (4),  it  is  evident  that  subtracting  a  nega- 
tive number  is  equivalent  to  adding  an  equal  positive  number. 

To  subtract,  therefore,  one  algebraic  number  from  another, 
change  the  sign  of  the  subtrahend,  and  then  add  the  subtra- 
hend to  the  minuend. 

Exercise  VI. 

1.  +25-(+16)=  3.  -   31-(+58)  = 

2.  -50 -(-25)=  4.   +107 -(-93)  = 

5.  Rome  was  ruled  by  emperors  from  B.C.  30  to  its  fall, 

A.D.  476.     How  long  did  the  empire  last? 

6.  The  continent  of  Europe  lies  between  36°  and  71°  north 

latitude,  and  between  12°  west  and  63°  east  longi- 
tude (from  Paris).  How  many  degrees  does  it  extend 
in  latitude,  and  how  many  in  longitude  ? 

Subtraction  of  Monomials. 

If  a  and  b  denote  the  absolute  values  of  any  two  num- 
bers, 1,  2,  3,  and  4  (§  51)  become : 

(1)  -\-a  —  {-\-b)  =  a  —  b.  (3)    —  a  —  {-\-b)  =  —  a  —  b. 

(2)  +a-(-5)  =  a  +  5.  (4)   —  a  — (- Z>)  =  — a  +  5. 

To  subtract,  therefore,  one  term  from  another,  change  the 
sign  of  the  term  to  be  subtracted  and  write  the  term,s  one 
after  tJie  other. 


22  ALGEBRA. 


EXEECISE   VII. 

1.  6x-(-4:x)=  6.  17a^-(-24aa;«)  = 

2.  -Sab-(+5ah)=  7.  5a^x- (-Sa'x)  = 

3.  3a^>2_(+i0a52)=  8.  -4:X7/-{-6x2j)  = 

4.  15mV  — (-7m2x2^=  9.  8  a^;  —  (- 3  ay)  = 

5.  — 7ay  — (-3ay)=  10.  2  abhj  ~  (+ aby)  = 

11.  9:i;2  +  (5^)-(+8rr2)  = 

12.  5a.^y-(-182;2^)_l_(_;i^0.^2^)=: 

13.  17a:^-(-aa;3)-(+24aa;^)  = 

14.  —  3  (2^  +  (2  wa;)  —  (—  4ma;)  = 

15.  3a -(+2^)- (-4c)  = 

Subtraction  op  Polynomials. 

53.  "When  one  polynomial  is  to  be  subtracted  from  an- 
otber,  place  its  terms  under  tbe  like  terms  of  tbe  other, 
change  the  signs  of  the  subtrahend,  and  add. 

From  4:0^  —  3x^7/—    x7/^-{-2y^ 

take  2a^~    x^y-}-5xy^  —  dy^ 

Change  the  signs  of  the  subtrahend  and  add  : 

Act^-Sx^y-    xy^  +  2f 
—  2a^+    x^i/—5xy^-{-Sy^ 

20^-2x^1/ -6xf +  5  f 

From  a^  x^  -{-  2a^  a^  —  4:ax* 

take  a''  +  4aV-3aV-4aa;* 


SUBTRACTION.  23 


In  the  last  example  we  have  conceived  the  signs  to  be 
changed  without  actually  changing  them.  The  beginner 
should  do  the  examples  by  both  methods  until  he  has  ac- 
quired sufficient  practice,  when  he  should  use  the  second 
method  only. 

Exercise  VIII. 

1.  Yrom6a  —  2b  —  ctakQ2a  —  2b  —  Sc. 

2.  ¥Tom^a-2b-{-Sctsi'ke2a-7b-c-b. 

3.  From  7:^2-807-1  take  5a.^  — 6a; +  3. 

4.  From4.2;*-3:r3_2^_7^_j_9 

tsikex*-2a^-2ar^~{-1x-9. 

5.  From  2a^  -  2ax  +  S a^  take  x^- ax +  aK 

6.  From  a;^  — 3^:3/- y^  +  ys;  — 22^ 

-     takex^  +  2x7/-\-bxz-S7/^-2z^ 

7.  Yiom  a^-Sa^  +  Sab^-b^ 

take  -  a^-\-Sa^b  -3ab^  +  b\ 

8.  From  x^  —  5xy-{-xz  —  y^-]-7yz-{-2z^ 

take  ar^  —  XT/ —  xz -\-  2yz  +  3  ^. 

9.  Yxom2ao?-{-^abx~Wx-^\2W 

take  aoi^  —  ^abx-^ba^  —  bb^x  —  o?. 

10.  YziQm^3?-nx^y^^xif-2f-bx^-\-xy-^'f^2 

izk^'^x?—nx^y-\-xf-f^9x^-xy-\-^f~^. 

11.  From  a*  — J*  take  A:a^b —  ^a^b'^ ^^ab^,  and  from  the 

result  take  2a>-^a^b-\-^ aH^ -^^ab^  -2b\ 

12.  From  oi^'i/  —  2>oi^if  -\-  ^lXi/  —  if  take  —0!^-{-2x^y  —  4:X7/ 

—  4y.     Add  the  same  two  expressions  together,  and 
subtract  the  former  result  from  the  latter. 

13.  From  aH'^-aHc-^ab^c  -  a'c'  +  ab(^  -(ob^c" 

take  2aV)c  -  bab''c  +  2ab(^ -U'c". 


24  ALGEBRA. 

14.  From   12a  +  35  -  5c- 2c?   take    lOa-b +  4:C  —  Sd, 

and  show  that  the  result  is  numerically  correct  when 
a  =  6,  b  =  4:,  (7=1,  d=b.  4- 

15.  What  number  must  be  added  to  a  to  make  h  ;  and  what 

number  must  be  taken  from  2a^  —  Ga^b-{-6 ab^  —  2b^ 
to  leave  a^-Ta^i-S^^? 

16.  FTom2x^-f-2x7j  +  zHakex^-y^-i-2xy-z\ 

17.  From  12ac  +  8cd—9  take  -  7 ac  —  9cd  +  8. 

18.  From  -6a^  +  2ab-Sc^  take  Aa^+Qab -4:^', 

19.  From  9xi/  —  Ax —  St/ -\-7  take  8x7/ —  2xi-Sy-\-6. 

20.  From  —  a^bc  —  aW  c  +  ab(f  —  abc 

take  a^  be  +  a5^  <?  —  abc^  +  «^^- 

21.  From  7:^2- 2:r  + 4  take  2^2 _^3:r-l. 

.22.    From  S a^ -j- 2 xt/  —  y^  take  —a^  —  Sx7/~{-  Sy^,  and  from 
the  remainder  take  Sx^-^-Axy  —  5yl 

23.  From  ax^  —  ^3/^  take  car^  —  c??/^. 

24.  From  ax  -\~  bx  -{-  by  -{-  cy  take  ax  —  bx~by-\-  cy. 

25.  From  5x2_^4^  _  4^_|_  3^  take  5:^2-3^  +  3^  +  2/2^ 

26.  From  o?  b^  +  12  abc  -dax^  take  4  aS^  —  6  ac:?;  +  Sa^x. 

27.  From  a^ -  2ab  +  d'-Sb^  take  2a2_  2a5  +  351 

28.  From  the  sum  of  the  first  four  of  the  following  expres- 

sions, a^  +  ^>2  _|_  ^2  _|_  ^2^  ^2_^J2^^^  a2_^_^^2_^-i^ 

a^-b^  +  c'  +  d^,  P-i-c^+d^-a^,  take  the  sum  of  the 
last  four. 

29.  From  2:^-2y^-z^  take  Sf  +  2xr^-z^  and  from  the 

remainder  take  3  s^  —  23/^  _  a^, 

30.  From  a^  —  2a^c-j-3ac^  take  the  sum  of  a^c  —  2a^-\-2ac^ 

and  a^  —  ac^  —  a^c. 


parentheses.  25 

Parentheses. 

54.  From  (§  52),  it  appears  that 

(1)  a  +  (+5)  =  «  +  ^». 

(2)  a  +  {~h)  =  a-h. 

(3)  a--{-\-h)  =  a-h. 

(4)  a-{r-b)  =  a  +  h. 

The  same  laws  respecting  the  removal  of  parentheses  hold 
true  whether  one  or  more  terms  are  inclosed.  Hence,  when 
an  expression  within  a  parenthesis  is  preceded  by  a  plus 
sign,  the  parenthesis  may  be  removed. 

When  an  expression  within  a  parenthesis  is  preceded  by 
a  minus  sign,  the  parenthesis  may  be  removed  if  the  sign 
of  every  term  within  the  'parenthesis  he  changed.     Thus  : 

(1)  a  +  (6  — <?)  =  «  + 5  — c. 

(2)  a  —  {}t  —  c)  =  a  —  h-\-c. 

55.  Expressions  may  occur  with  more  than  one  paren- 
thesis. These  parentheses  may  be  removed  in  succession, 
by  removing  ^rs^,  the  innermost  parenthesis ;  next,  the  inner- 
most of  all  that  remain,  and  so  on.     Thus : 

(1)    a-\h-{c-d)\ 
=  a  —  \b  —  c?  +  c?j, 
=  a  —  b-i-c  —  d. 


(2)    a-[b-\c  +  (d-e-f)\] 
=  a-[b-\c  +  (d-e+f)\l 
=  a-[b-\e  +  d-e+f]l 
=  a—[b  —  c  —  d-{-e  — /], 

=  a-b-j-c  +  d~e+f 


26  ALGEBRA. 

Exercise  IX. 

Simplify  the  following  expressions  by  removing  tlie  paren- 
theses and  combining  like  terms. 

1.  {a  +  h)-{-{b-^c)-{a  +  c). 

2.  (2a-h-c)-{a~2b-\-c). 

3.  {2x-y)-{2y-z)-(2z-x). 

4.  {a  —  x  —  y)  —  (h  —  x  +  y)-\-{c  +  2y). 

5.  (2x  -  y  -\-^z)  -{-  {-  X  -  y  —  4:z)  -  (?>x—2y  -  i). 

6.  (2>a-h-^lc)-{2a  +  U)-{bh-^c)  +  Q^c-a). 

7.  1-  (I-  a)  +  (I-  a  +  a^)  -  (I-  a-}-  o?  -  a% 

8.  a-\2h-{^c  +  2h)-a\. 

9.  2a-\h-{a-2h)]. 

10.  2>a-\h  +  {2o.-h)-{a-h)\. 

11.  7a-[3a-J4a-(5a-2a)|]. 

12.  2x  +  {y-'^z)-\{^x-2y)  +  z]-\-bx-{4:y-^z). 

13.  K3a-25)  +  (4c-a)i-Ja-(25-3a)-t?} 

+  Ja  — (Z>  — 5(?  — a)J. 

■  14.   a-[2a+(3a-4a)]-5a-J6a-[(7a  +  8a)-9a]f. 

15.  2a-(35  +  2c)-[55-(6c-65)  +  5(? 

-\2a-{c  +  2h)\l 

16.  a-[25  +  !3c-3a-(a  +  5)|+f2a-(5  +  c)|]. 

17.  16 -:r- [7^ -58a;- (9a;- 3^ -6^)5]. 


18.  2a-[U  +  (2h-c)-4.c  +  \2a-{U-c-2h)]\ 

19.  a-  [25  +  53c -  3a -  (a  +  5) j  +  2a -  (5  +  3c)]. 

20.  a-[bb  -{a-  i^c  -U)  +  2c  -  (a-2h  -  c)\l  j 


PARENTHESES.  27 


56.  The  rules  for  introducing  parentheses  follow  directly 
from  the  rules  for  removing  them : 

1.  Any  number  of  terms  of  an  expression  may  be  put 
within  a  parenthesis,  and  the  sign  plus  placed  before  the 
whole. 

2.  Any  number  of  terms  of  an  expression  may  be  put 
within  a  parenthesis,  and  the  sign  minus  placed  before  the 
whole ;  provided  the  sign  of  every  term,  within  the  paren- 
thesis he  changed. 

It  is  usual  to  prefix  to  the  parenthesis  the  sign  of  the 
first  term  that  is  to  be  inclosed  within  it. 

Exercise  X. 

Express  in  binomials,  and  also  in  trinomials : 

1.  2a-35-4(?  +  c?+3e-2/. 

2.  a  —  2x-^^y  —  ^z  —  2b'{-c. 

3.  a«  +  3a*-2a3-4a2  +  a_l. 

4.  -3a-25+.2c-5c?-e-2/. 

5.  ax  —  hy  —  cz  —  hx -{- cy -{-  az.  * 

6.  2r'-3a;V  +  4:r3y'-5a:Y  +  V-2/.    ' 

7.  Express  each  of  the  above  in  trinomials,  having  the 

last  two  terms  inclosed  by  inner  parentheses. 

Collect  in  parentheses  the  coeflicients  of  x,  y,  z  in 

8.  2ax  —  ^ay-\-^hz  —  4:hx  —  2cx  —  2>cy. 

9.  ax  —  hx-\-2ay-[-2>y-\-^az  —  ^bz  —  2z. 

10.  ax  —  2hy  -\-  b cz  —  4:hx  —  ^ cy  -\-  az  —  2cx  —  ay  -\-  4ihz. 

11.  12ax  +I2ay  -\-  A:hy  -\2hz  —  Ibex  -^  Q>cy  -\-^cz. 

12.  2ax  —  Sby  —  7 cz  —  2bx i- 2cx -i- Scz  —  2cx  —  cy  —  oz. 


CHAPTER  III. 

Multiplication  of  Algebraic  Numbers. 

57.  The  operation  of  finding  the  sum  of,  3  numbers,  each 
equal  to  5,  is  symbolized  by  the  expression,  3x5  =  15, 
read,  "three  times  five  is  equal  to  fifteen";  or,  by  the 
expression  5  X  3  =  15,  read,  "  five  multiplied  by  three  is 
equal  to  fifteen." 

58.  With  reference  to  this  operation,  this  sum  is  called 
the  product ;  one  of  the  equal  numbers  is  called  the  multi- 
plicand ;  and  the  number  which  shows  how  many  times  the 
multiplicand  is  to  be  taken  is  called  the  multiplier. 

59.  The  multiplier  means  so  many  times.  The  multipli- 
cand can  be  a  positive  or  a  negative  number  ;  but  the  mul- 
tiplier can  only  mean  that  the  multiplicand  is  taken  80 
many  times  to  he  added,  or  so  many  times  to  he  subtracted. 

60.  If  we  have  to  multiply  867  by  98,  we  may  put  the 
multiplier  in  the  form  100  —  2.  The  100  will  mean  that 
the  multiplicand  is  taken  100  times  to  he  added;  the  —  2 
will  mean  that  the  multiplicand  is  taken  twice  to  he  sub- 
tracted. 

In  general,  a  multiplier  with  +  before  it,  expressed  or 
understood,  means  that  the  multiplicand  is  taken  so  many 
times  to  he  added;  and  a  multiplier  with  —  before  it  means 
that  the  multiplicand  is  taken  so  many  times  to  he  sub- 
tracted.    Thus, 


MULTIPLICATION.  29 


(1)  +3x(+5)  =  (+5)  +  (+5)  +  (+5),or(+15). 

(2)  +3x(-5)  =  (-5)  +  (-5)  +  (-5),or(-15). 

(3)  -  3  X  (+  5)  =  -  (+  5)  -  (+  5)  -  (+.5),  or  (-  15). 

(4)  -3  x(- 5)  =  -(-5) -(-5) -(-5).  or  (+15). 


1 


From  these  four  cases  it  follows,  that,  in  multiplying  two 
numbers  together, 

61.    lATce  signs  produce  plus  ;  unlike  signs,  minus. 


Exercise  XI. 

1.  -17x8=  4.  -18x-5  = 

2.  -12.8x25=  5.   43  X -6  = 

3.  -3.29x5.49=  6.   457x100  = 

7.  (-358-417)X-79  = 

8.  (7.512 - ;-  2.894S)  x  (-  6.037  +  ;i3.963|)  = 

62.  The  product  of  more  than  two  factors,  each  preceded 
by  — ,  will  be  positive  or  negative,  according  as  the  number 
of  such  factors  is  even  or  odd.     Thus, 

-2x-3x-4=   +6x-4=   -24. 
-2x-3x-4x-5  =  -24x-5  =  +  120. 

9.  13x8x  — 7  = 

10.  -  38  X  9  X  -  6  = 

11.  —  20.9  X- 1.1  X8  = 

12.  -  78.3  X  -  0.57  x  + 1.38  X- 27.9  = 

13.  -  2.906  X  -  2.076  x  -  1.49  X  0.89  = 


30  ALGEBRA. 


Multiplication  of  Monomials. 

63.  The  product  of  numerical  factors  is  a  new  number 
in  which  no  trace  of  the  original  factors  is  found.  Thus, 
4  X  9  =  36.  But  the  product  of  literal  factors  can  only  be 
expressed  by  writing  them  one  after  the  other.  Thus,  the 
product  of  a  and  h  is  expressed  by  ab ;  the  product  of  ah 
and  cd  is  expressed  by  ahcd. 

64.  If  we  have  to  multiply  5  a  by  —4  Z),  the  factors  will 
give  the  same  result  in  whatever  order  they  are  taken. 
Thus,  5aX— 4S  =  5x  — 4XaX&  =  — 20x  ab  =  —  20ab. 

65.  Hence,  to  find  the  product  of  monomials,  annsx  the 
literal  factors  to  the  'product  of  the  numerical  factors. 

66.  a^  X  a?^=aaX  aaa 
o?  X  a^  X  a^  =  aaX  aaa  X  aaaa  =  aaaaaaaaa  =  c^. 

It  is  evident  that  the  exponent  of  the  product  is  equal  to 
the  sum  of  the  exponents  of  the  factors.     Hence, 

67.  The  product  of  two  or  more  powers  of  any  number  is 
that  number  with  an  exponent  equal  to  the  sum  of  the  expo- 
nents of  the  factors. 

Exercise  XII. 

1.  +aX+J  =  +  a5.  6.  —  3^X  8m  =  — 24j9m. 

%  -\-aX  —  b  =  —  ab.  7.  Sa^  x  -  a^^  — 3a^ 

Z.  —aX  +  b  =  —  ab.  8.  —  3a  X  2a''  =  — 6a^ 

4.  —  aX  — 2>  =  +  a5.  9.  6aX— 2a  = 

5.  7ax55  =  35a5.  10.  5mnX9m  = 


MULTIPLICATION.  81 


11.  3a;rX— 453/=  15.    5a'"X  — Sa'*^ 

12.  —8cmXdn=  16.    3aV  X  7a^x*  = 

13.  —  7aJx2ac=  17.    7aX— 4^X  — 8c  = 

14.  5m^xxSma^=  la    8aP  X  Sac  X  -  4:C^  = 

19.  27a5x  — 39wpX  18ap  = 

20.  6aPy^x2Pfx-5a^y  = 

21.  Ym^^r  xSmar^X  — 2m2'  = 

22.  -Spq^x6p^qX8p^^  = 

23.  2a2y?z3a;4  x  3am* a;^  x  4a3m:r2 ^ 

24.  6^:2^23  x_9^y5^X_3^4y2^ 

25.  3aa:X  2am  X  —  4m2:  X /^^  = 

26.  7am2  X  3  i^n^  X  -4a^'  X  aHn  X~2b^nX~mn^  = 

Of  Polynomials  by  Monomials. 

68.    If  we  have  to  multiply  a-}-h  by  n,  that  is,  to  take 
(a-{-h)n  times  to  be  added,  we  have, 

(a+b)Xn  =  (a  +  h)  +  (a  +  h)  +  (a  +  b)....n  times, 

~a-{-a-\ra.  .  .71  times -{-b-\-h-{-b n times, 

=  aXn-i-bXn, 
=  an-{-  bn. 

As  it  is  immaterial  in  what  order  the  factors  are  taken, 

nX{a  +  b)  —  an-\-  bn. 
In  like  manner, 

(a  +  5  +  c)  X  w  =  an  +  5w  +  en, 
or,  n{a  +  b-]-c)  =  an-\'bn  +  cn. 


32  ALGEBRA. 


Hence,  to  multiply  a  polynomial  by  a  monomial, 

69.    Multiply  each  terin  of  the  polynomial  hy  the  monomial, 
and  add  the  partial  products. 


Exercise  XIII. 

1.  {i6a-bh)X^c  =  lSac-lbhc. 

2.  (2  +  3a-4a2_5a3)6a2  =  12a2+18a3-24a*-30a^ 

3.  ba{?>h  +  4:C-d)  =  lbab-{-20ac-had. 

4.  —3  ax{—hy~2cz-{-b)-=Z  ahxy  +  6  acxz  — 15  ax. 

5.  (4a2_3Z^)x3a&  = 

6.  (8a2-9«5)x3a2  = 

7.  {^x'-^y'  +  bz'')x2a?y  = 

8.  {a^x  —  ba^x^  +  ao(^  +  2x*)  X  ac(Py  = 

9.  {:-%a^-\-?>a^l/-4:a'l/--l/)X-'^ah^  = 

10.  (3:^-^-2^y-7V  +  2/')x-S^y  = 

11.  (-4V  +  5:i^2/  +  8:r3)x-3a;2^^ 

12.  (-3  +  2aJ  +  a2J2^)><_^4^ 

13.  (-  z  -  2a;z2  +  5^y22 -  6^/ _|_  3^3^2^>j  ><  _  3^^^  ^ 

Of  Polynomials  by  Polynomials. 

70.    If  we  have  a  +  J  +  c  to  be  multiplied  by  m  +  w  +^, 
we  may  represent  the  multiplicand  a  +  5  +  ^  by  M.     Then, 

Jf(m  +  71 +^)  =  Jf  X  m  +  J[f  X  w  +  if  X  ^. 

If  now  we  substitute  for  J/"  its  value, 

{a  -\-h  -\-  c)  {m  -\-  n  -{-p)  =  {a  +  h-{-c)Xm 
+  (a  +  5  +  c)  X  72 
+  (a  +  5  +  c)Xp; 


MULTIPLICATION.  33 


or,        (a -{- b -{- c)  (7?i -}- n  +/>)  =  ani  +  bm  +  cm 

-\-an  -]-bn  -\-en 
+  ap  +bp  +cp. 

That  is,  to  find  tlie  product  of  two  polynomials, 

71.  Multiply  the  multiplicand  by  each  term  of  the  Tnultiplier 
and  add  the  partial  products;  or^  m^ultiply  each  term  of  one 
factor  by  each  term  of  the  other  and  add  the  partial  products. 

72.  In  multiplying  polynomials,  it  is  a  convenient  ar- 
rangement to  write  the  multiplier  under  the  multiplicand, 
and  place  like  terms  of  the  partial  products  in  columns. 

Thus:  (1)      5a  -   6  5 

3a  -   4   5 

15a2-18a6 

-20a^>  +  245^ 
15a*-38a6  +  242>* 

(2)  Multiply4a;  +  3  +  5a:2_6^l3y4_6^_5^^ 
Arrange  both  multiplicand  and  multiplier  according  to 

the  ascending  powers  of  x. 

3+   4a;+    ^^-   6r» 
4—    hx—    6a;^ 

12  +  16ar  +  20r»-24a;« 

-  15a;  -  20a:2  __  25a:3  _|_  30^4 

-18a;^-24:g^-302:^  +  36:r« 

12+      x-V^x^-^iZx"  +36:zr* 

(3)  Multiply  l  +  2a:  +  a;*-3a,-2bya;3_2_2a;. 
Arrange  according  to  the  descending  powers  of  x. 

a;*-3a:2_|_2a;  +1 
a:^  — 2a:  —2 

a;'-3ar^  +  2a-*+    3? 

-2a^  +Qix^-^a^~2x 

-2a;^  +6a'^-4ar-2 

x'-b:^^  -\-la?-\'2x'~Q>x-2. 


34  ALGEBRA. 


(4)    Multiply  a^-^h^-\-(^~ah  —  hc~achja  +  h  +  c. 
Arrange  according  to  descending  powers  of  a. 

a  -\-     h-\-     c 

+  aH  -ah^-    ahc  J^  h^ -  y^ c -\- he" 

-\-a^c —    abc  —  ac^          ■\-h'^c  —  hc^-\-(? 

a^  -Zabo  +h^  +c^ 

The  student  should  observe  that,  with  a  view  to  bringing 
like  terms  of  the  partial  products  in  columns,  the  terms  of 
the  multiplicand  and  multiplier  are  arranged  in  the  same 
order. 

In  order  to  test  the  accuracy  of  the  work,  interchange 
the  multiplicand  and  multiplier.  The  result  should  be  the 
same  in  both  operations. 

Exercise  XIV. 
Multiply  : 

1.    aP-^hj  aP  +  b.  3.    a^  +  a"  x" -\- x""  hj  a"  -  x^ . 

%   y  — 6by?/+13.  4.   x^ -{-xy  ^y^'^y  x  —  y. 

5.  2^x--y\>jx-\-'^y. 

6.  2a;^  +  4^  +  8^  +  16by  3:r-6. 

7.   x^-\-3?-\-x  —  \\>Y  x  —  \.     8.   oc^  —  ?>ax\)j  x^Za. 
9.    2^2  +  3aZ>-a2by-5&  +  7a. 

10.  2a  +  Z»bya  +  25.  12.    a^ -ah '\-h'^\ij  a-^l. 

11.  ce-\-ab-\-W\ij  a-l.         13.   2a5  — 5Z»2  by  3a2_4aZ'. 

14.  -a3  +  2a25-53by4a2  +  8a5. 

15.  a^-\-ah-\-y^^ja^-ah-\-l\ 


MULTIPLICATION.  35 


16.  a^-^an-^Zah^-Vhy  a:'-2ah-\-h\ 

17.  a:  +  2y-32by  :r-2y  +  3z. 

18.  2:i^  +  2>xy  +  ^7fhj^x'-^xij-^yz. 

19.  o(^ -^  xy -\- if  hj  x^ -\- xz  ^  z^. 

20.  a^  -{-  V^  -\-  c^  —  ab  —  ac  —  he  \>Y  a  -\-h  -{•  c, 

21.  x^  — ccy-\-'f-\-x-\-y-\-l\)jx-\-y  — \. 

Arrange  the  multiplicand  and  multiplier  according  to 
the  descending  powers  of  a  common  letter,  and  multiply  : 

*       22.  bx  +  ^c^-\-a^-24:hY  x'+ll-^x. 

23.  x^-\-llx-4:a^-2^hy  x^-{-b-{-4:X. 

24.  x'^-\-o^  —  Ax-ll-^2x^hyx^~2x-{-^, 

25.  -bx^-a^-x  +  a^  +  l^s^hy  x^~2-2x. 
j^l26.  ^x-\-3?-2x^-A:hY2x  +  4:a?  +  ^a?+l. 

27.    ba''-\-2o?h''-\-al/ -Sa^h  by  6a^b - 2ah^+Sa'P  +  h\ 

/    28.   4aV-32a/-8aV  +  l^«ybyaV  +  4aV  +  4aV- 

V  29.    3 m'  +  3 n^  +  9 T^in^  +  9 m^n  by  6 7n^7i^  -  2 mri* 

—  67u^n^-{-27n*n. 

V  30.    6a'h  +  Sa'L*-2ab'  +  h^hY4:a*-2ah^-Sb\ 

Find  the  products  of : 

31.  x  —  S,  x—1,  x  +  l,  and  x-\-S. 

32.  x^-x+l,  a^  +  x-{-l,  and  :i'^-.x^-f  1. 

33.  a^-{-ab  +  h\  a"  -  ab  +  ^'^  and  a*  -  cc"  b^  +  Z.^ 

>/  ^34.    4^^3_4^^2^_l_^^2^  4a2  +  3a5  +  i2^  and  2a^b  +  b\ 
35.    .2:  + a,  a; +  2 a,  a;  — 3a,  a:  — 4a,  and  :r  +  5a. 
L36.    9a2  +  Z>2,  27a^-Z*«,27a3  +  ^',  and  81a^-9a^P+b\ 


36  ALGEBRA. 


37.  From  the  product  oiy^  —  2yz—    z^  and  y^-\-2yz—    ^ 

take  the  product  of  y^  —    yz  —  2z^  and  y^  +    2/2;  —  2 z^. 

38.  Find  the  dividend  when  the  divisor  —  3  a^  —  a5  —  3  5^, 

the  quotient  =a^b  —  2b^,  the  remainder  ==  — 2a^'* 

The  multiplication  of  polynomials  may  be  indicated  by 
inclosing  each  in  a  parenthesis  and  writing  them  one  after 
the  other.  When  the  operations  indicated  are  actually  per- 
formed, the  expression  is  said  to  be  simplified. 

Simplify : 

39.  (a  +  5 - c)(a  +  c~h){b -\-c-a){a^h  +  c). 

40.  {a  +  h){b-\-c)-{c  +  d){d-\-a)-{a-\-c){h-d). 

41.  {a^b-\-c^df-\-{a-b-c-\-df 

-\-la-b  +  c  -  df  -^  (a-\-b  ~  c  -  d)\ 

42.  {a-\-b-\-cf—a{b-\-c  —  d)  —  b{a-\-c—b)  —  c{a-\-b—c), 

43.  {a  —  b)x-{b-c)a  —  \(b—x){b  —  a)  —  {b  —  c){b-\-c)\. 

44.  {7rl■\•r^)rl^  —  \{rrl  —  nf--{n  —  7n)n\. 

45.  {a-b-\-cf-{a{c-~a-b)~[b{a-\-b+c) 

—  c{a  —  h  —  cy\\. 

46.  {p'+q^r-ip  +  qXplr-ql-qlr-pl). 

47.  {^:^f-^y'){x'-f)-\2>xy-2f\\^x{x'  +  f) 

-2y{f  +  2>xy-x')\y. 

48.  a2-{2a5-[-(a+S^>-cJ)(a-J^-^D  +  2«^] 

4Q.   \ac-{a-b)(b  +  c)\-b\b-{a-c)\. 

50.   5  |(a  —  b)x  —  cy\~  2  \a  (x  ~  y)  —  bx] 

~\2>ax-{bc-2a)y\. 

61.   (a;-l)(a;-2)-3a;(a;  +  3)  +  2K^  +  2X^+l)~3i. 


MULTIPLICATION.  37 


52.    \(2a^hf-\-{a-2bf\x\{^a-2bf-(2a-Uf\. 
58.   4(a-35)(a  +  35)-2(a-66)^-2(a2  +  652). 

54.  a?{a?-^yJ~2:^f{x-\-yXx-y)-{:!?-f)\ 

55.  16  {a?  +  b%a^  -  5^  -  (2  a  -  3)(2  a  +  3)(4  a*  +  9) 

+  (2^»-3)(26  +  3)(4^>H9). 


73.  There  are  some  examples  in  multiplication  which 
occur  so  often  in  algebraical  operations  that  they  should 
be  carefully  noticed  and  remembered.  The  three  which 
follow  are  of  great  importance: 


a  -{-      h            (2)    a  -      b 

(3)    a  +   5 

a  •{-      h                    a  —      b 

a-  b 

a^+     «^                     0?—    ah 

c^  +  ab 

ab-\-b^                -    ab-\-b' 

-ah- 

-b' 

a^  +  2ab  +  b^             o?-2ab  +  b' 

a^ 

-b' 

From  (1)  we  have  (a  +  by  =  a^-{-2ab-i-  b\     That  is, 

74.  The  square  of  the  sum  of  two  numbers  is  equal  to  the 
sum,  of  their  squares  +  twice  their  product. 

From  (2)  we  have  (a -bf  =  a^-2ab  +  b\     That  is, 

75.  The  square  of  the  difference  of  two  numbers  is  equ/xl 
to  the  sum  of  their  squares  —  twice  their  product. 

From  (3)  we  have  (a  +  b){a -b)  =  a^- b\     That  is, 

76.  The  product  of  the  sum  and  difference  of  two  num- 
bers is  equal  to  the  difference  of  their  sqicares. 

77.  A  general  truth  expressed   by  symbols  is  called  a 
formula. 


38  ALGEBRA. 


.  78.    By  using  the  double  sign  -\z,  read  plus  or  minus,  we 
may  represent  (1)  and  (2)  by  a  single  formula ;  thus, 

{a-\,by  =  a^-k2ah  +  b^', 

in  which  expression  the  upper  signs  correspond  with  one 
another,  and  the  lower  with  one  another. 

By  remembering  these  formulas  the  square  of  any  bino- 
mial, or  the  product  of  the  sum  and  difference  of  any  two 
iiumbers,  may  be  written  by  inspection ;  thus : 


Exercise  XV. 

1.  (127)2  -  (123)2  ^  (^2.27  + 123)(127  - 123) 

=  250x4  =  1000. 

2.  (29)2  =  (30 -1)2  =  900- 60  + 1  =  841. 

3.  (53)2  ^  (50  _|.  3)2  ^  2500  +  300  +  9  =  2809. 

5.  (2a2a;  _  5^y)2  3^  4^4^ _  20a^x^7/  +  25a;V- 

6.  (Sab''c  +  2a^c'XSab'c-2a'e')  =  9a'b'c'-4:a'c\ 

7.    (x  +  7/y=  15.    (ab  +  cdf 


\2 


8.  (y-z)2=  16.  (3m7z-4)2  = 

9.  (2a: +  1)2=  17.  (12  +  5^)2  = 

10.  (2a  +  55)2=  18.  (4xf-yzy  = 

11.  (1-^)2=  19.  (Sabc-bcdy  = 

12.  (3  aa:- 4:^2)2=-  20.  (4;r«- V)'  = 

13.  (l-7a)2=  21.  (^  +  y)(^-y)  = 

14.  (5^:2/ +  2)2=  22.  (2a-^b)(2a-b)  = 


MULTIPLICATION.  39 


23.  (S-x)(S  +  x)  = 

24.  (Sah  +  2h')(^ah-2h')  = 

25.  (ia^-Sy')(^3^  +  Sf)  = 

26.  (a'x'-hi/'Xa^x'+by')  = 

27.  (6a;y-5y2)(6:ry  +  52/^)  = 

28.  (4:?^-l)(4:r«+l)  = 

29.  {1  +  Sab')(l-Sah^)  = 

30.  (aa;  +  63/)(aa:-5y)(aV+5V)  = 

79.    Also  the  square  of  a  trinomial  should  be  carefully- 
noticed. 


a  +      ^  + 

c 

a+      ^  + 

c 

a2+    a6  + 

ac 

ah 

+  ^2+    he 

ae          -\-    hc-{-(^ 

a^  +  2a6  +  2ac  +  62  +  26c  +  c2, 
=  a2  +  52  +  c2  +  2a6  +  2ac  +  25c. 

It  is  evident  that  this  result  is  composed  of  two  sets  of 
numbers : 

I.   The  squares  of  a,  h,  and  e ; 
II.   Twice  the  products  of  a,  6,  and  c  taken  two  and  two. 

Again, 


a  — 

h- 

c 

«  — 

h- 

c 

a^- 

ah- 

ac 

— 

ah 

+  52+    he 
ac         -\-    he-[-c^ 

a^-2ah-2ae^h^  +  2he+.c' 
a^-\-h^-\-c'-2ab-2ac  +  2hc. 


40  ALGEBRA. 


The  law  of  formation  is  the  same  as  before : 

I.   The  squares  of  a,  h,  and  c ; 
II.   Twice  the  products  oi  a,h,  and  c  taken  two  and  two. 

The  sign  of  each  double  product  is  +  or  —  according  as 
the  signs  of  the  factors  composing  it  are  like  or  unlike. 

The  same  law  holds  good  for  the  square  of  expressions 
containing  more  than  three  terms,  and  may  be  stated  thus : 

80.  To  the  sum  of  the  squares  of  each  term  add  twice  the 
product  of  each  term  into  each  of  the  terms  that  follow  it. 

By  remembering  this  formula,  the  square  of  any  polyno- 
mial may  be  written  by  inspection  ;  thus  : 

Exercise  XVI. 

1.  {x-{-y-{-zf=  9.  {a^^h^-{-(^y  = 

2.  (x-y-\-zf=  10.  {p(^-f-^f  = 

3.  (m  +  n-p-qf^  11.  {x  +  2y  —  2>zf  = 

4.  (ar^  +  2^-3)2=  12.  (a:2_2y2_|-5/)2^ 

5.  {s^-^x  +  lf^  13.  {x'  +  2x-2f  = 

6.  (2^-7a;  +  9)2=  14.  {x'-bx-^1f  = 

7.  {x'  +  f~zy=  15.  {2x'-2>x~^f  = 

8.  (x''-4::x^'/-\-yy=  16.  {x-{-2y  +  ^zf  = 

81.  Likewise,  the  product  of  two  binomials  of  the  form 
a:  +  a,  x-{-h  should  be  carefully  noticed  and  remembered. 

(1)     ^+5  (2)     X  -b 
X +3  X —3 

0(?-{-bx  a?  —  f>x 

3a;+15  -3a:  +  15 


MULTIPLICATION.  41 


(3)     :r  +  5  (4)     X  -b 

X -3  X +3 

o(^-\-bx  x^  —  bx 

-3a;-15  +3a;-15 

a^-{-2x-lb  a?-2x-lb 

It  will  be  observed  that : 

I.  In  all  the  results  the  first  term  is  3?  and  the  last  term 
is  the  product  of  5  and  3. 

II.  From  (1)  and  (2),  when  the  second  terms  of  the  bino- 
mials have  like  signs,  the  product  has 

the  last  term  positive  ; 

the  coefficient  of  the  middle  term  =  the  sum  of  3  and  5  ; 
the  sign  of  the  middle  term  is  the  same  as  that  of  the 
3  and  5. 

III.  From  (3)  and  (4),  when  the  second  terms  of  the  bi- 
nomials have  unlike  signs,  the  product  has 

the  last  term  negative ; 

the  coefficient  of  the  middle  term  =  the  difference  of 
3  and  5; 

the  sign  of  the  middle  term  is  that  of  the  greater  of  the 
two  numbers. 

82.    These  results  may  be   deduced   from   the   general 

formula, 

(x-\-a){x  +  b)  =  x^  +  {a^b)x  +  ab, 

by  supposing  for  (1)  a  and  h  both  positive ; 

(2)  a  and  h  both  negative ; 

(3)  a  positive,  b  negative,  and  a  >  6  ; 

(4)  a  negative,  b  positive,  and  a>b. 

By  remembering  this  formula  the  product  of  two  bino- 
mials may  be  written  by  inspection ;  thus : 


42 

ALGEBRA. 

Exercise  XVII. 

1. 

(^  +  2X^  +  3)-                11. 

(x-c)(x-d)  = 

2. 

(x  +  l)(x  +  b)=                12. 

(^-4yX^  +  y)  = 

3. 

(x  -  S)(x  -  6)  =                 13. 

(a-2b)(a-5b)  = 

4. 

(57-8X^-1)=                 14. 

(x^  +  2f)ix^  +  f)  = 

5. 

(x-8)(x+l)=                15. 

{f~Zxy){:^Arxy)  = 

6. 

(x  -  2)(x  +  5)  =                 16. 

iax  -  9)(ax  +  6)  = 

7. 

(x-S)(x+1)=                 17. 

{x  +  aXx-h)  = 

8. 

(a; -2X^^-4)-                 18. 

{x~ll){x-\-4.)  = 

9. 

(:.+ 1X^+11)=               19. 

(^+12X^-11)  = 

10. 

(x-2a)(x  +  ^a)=           20. 

(^-10X^-5)== 

83.    The  second,  third,  and  fourth  powers  of  a  +  5  are 
found  in  the  following  manner : 


a  + 

a  + 


c?-\-    ah 

^ ah  +  y" 

a  +h 


aH  +  2aJy'-\-h^ 
{a  +  hf  =  a^  -\-?>aH  ^-^aV"  -{-¥ 
a  +h 


an  +  Sa^P-i-Sah^  +  b* 
(a  +  by  =  a^  +  4:an-j-6a'b^  +  4.ab^-^b* 


MULTIPLICATION,  43 


From  these  results  it  will  be  observed  that : 

I.  The  number  of  terms  is  greater  by  one  than  the  ex- 
ponent of  the  power  to  which  the  binomial  is  raised. 

II.  In  the  first  term,  the  exponent  of  a  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised ; 
and  it  decreases  by  one  in  each  succeeding  term. 

III.  h  appears  in  the  second  term  with  1  for  an  expo- 
nent, and  its  exponent  increases  by  one  in  each  succeeding 
term. 

IV.  The  coefficient  of  the  first  term  is  1. 

V.    The  coefficient  of  the  second  term  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised. 

VI.  The  coefficient  of  each  succeeding  term  is  found 
from  the  next  preceding  term  by  multiplying  its  coefficient 
by  the  exponent  of  a,  and  dividing  the  product  by  a  num- 
ber greater  by  one  than  the  exponent  of  b. 

84.  If  b  be  negative,  the  terms  in  which  the  odd  powers 
of  b  occur  are  negative.     Thus : 

{a-by  =  a^  -  ^an  +  ^a?¥  -  4.ab^  -\-b\ 


Exercise  XVIII. 
Write  by  inspection  the  results : 

1.  {x-{-af=               5.    {x-{-af=  9.  (a:-fy)*  = 

2.  {x~af-=               6.    {x-ay=  10.  {x-yf  = 

3.  {x-{-lf=               7.    {x  +  iy=  11.  {x-{-Vf  = 

4.  (x-Vf^               8.    (.^-1)*=  12.  (.r-1)*- 


CHAPTER  IV. 

Division. 

85.  Division  is  the  operation  by  wHch,  when  a  product 
and  one  of  its  factors  are  given,  the  other  factor  is  deter- 
mined. 

86.  "With  reference  to  this  operation  the  product  is  called 
the  dividend ;  the  given  factor  the  divisor ;  and  the  required 
factor  the  quotient. 

87.  The  operation  of  division  is  indicated  by  the  sign  -v-; 

by  the  colon  : ,  or  by  writing  the  dividend  over  the  divisor 

12 
with  a  line  drawn  between  them.    Thus,  12  -^  4, 12  :  4,  — , 

each  means  that  12  is  to  be  divided  by  4. 

88.  +  12  divided  by  +  4  gives  the  quotient  +  3  ;  since 
only  a  positive  number,  +  3,  when  multiplied  by  +  4,  can 
give  the  positive  product,  +  12.  §  61. 

-f- 12  divided  by  —  4  gives  the  quotient  —  3  ;  since  only 
a  negative  number,  —  3,  when  multiplied  by  —  4,  can  give 
the  positive  product,  +12.  §  61. 

—  12  divided  by  +  4  gives  the  quotient  —  3  ;  since  only 
a  negative  number,  —  3,  when  multiplied  by  +  4,  can  give 
the  negative  product  —  12.  §  61. 

—  12  divided  by  —  4  gives  the  quotient  +  3  ;  since  only 
a  positive  number,  +  3,  when  multiplied  by  —  4,  can  give 
the  negative  product,  —  12.  §  61. 


DIVISION.  45 


(1)    ±f  =  +  3.  (3)    ^  =  -3. 


(iS)    ^  =  -3.  (4)    ^  =  +  3. 


From  (1)  and  (4)  it  follows  that 

89.  The   quotient  is  positive  when   the   dividend   and 
divisor  have  like  signs. 

From  (2)  and  (3)  it  follows  that 

The  quotient  is  negative  when  the  dividend  and  divisor 
have  unlike  signs. 

90.  The  absolute  value  of  the  quotient  is  equal  to  the 
quotient  of  the  absolute  values  of  the  dividend  and  divisor. 


Exercise  XIX. 


,     +264_  «    +3840_ 


^    106.33 


+  4  -30  -4.9 


^   -4648^  ^    -2568  «    -42.435 


8  +12  +34.5 


-264^  -7.1560 

+  24  +324 


-  3670  ^         - 1 


-85  -3.14159 


+  6.8503  _  J2,     -31831 


-61  -31.4159 


46  ALGEBRA. 


Division  of  Monomials. 

91.  If  we  have  to  divide  ahc  by  he,  aabx  by  ahy,  12ahc 
by  —4:ab,  we  write  them  as  follows : 

abc aabx ax  12  ahc q 

he         '  aby        y  —Aab 

Hence,  to  divide  one  monomial  by  another, 

92.  Write  the  dividend  over  the  divisor  with  a  line  between 
thcni ;  if  the  expressions  have  common  factors,  remove  the 
comvion  factors. 

If  we  have  to  divide  a^  by  a^,  a^  by  a^,  a^  by  a,  we  write 
them  as  follows : 


cc" 

aaaaa 

-,3 

a^ 

aa 

-  aaa 

—  a, 

a' 

aaaaaa 

=  a^ 

a'' 

— 

aaaa 

=  aa 

a' 

aaaa 

^3 

— 

—r- 

aaa  — 

-  a  . 

93.  That  is,  if  a  power  of  a  number  be  divided  by  a 
lower  power  of  the  same  number,  the  quotient  is  that  power 
of  the  number  whose  exponent  is  equal  to  the  exponent  of  the 
dividend  —  that  of  the  divisor. 

Again, 


a^         aa           11 

a^  ~  aaaaa  ~  aaa  ~  o? ' 

0?        aaa         1        1 

a^  ~  aaaaa  ~  aa~  a^' 

a^          aaaa             1 

1 

a^  ~~  aoAiaaaaa  ~  aaaa  ~ 

a^ 

DIVISION.  47 


94.  That  is,  if  any  power  of  a  number  be  divided  by  a 
higher  power  of  the  same  number,  the  quotient  is  expressed 
hy  1  divided  hy  the  number  with  an  expon£nt  equal  to  the 
exponent  of  the  divisor  —  that  of  the  dividend. 

Exercise  XX. 


^  T^-"^^-  ^'    2bc 


2.  ±^  =  -5.  a     ^ 

—  a 


-3? 


a  :-^  =  -h.         9.  -'^'^^^'^ 


-{•a  —2m  mp^x^ 


—  ah_  ■   7  ^^    Seabed _  «    —  51  g^c??/^  _ 

—  a  '  '      bbd  Sbdy 

6mx_  --     ahx__  ^   225 m^y__ 

2a;  *  5a5y  '    25?ny^ 

19  i^m^_  21.  =l-?^'i'^^-  = 
*  ba^m^x  '     ~a*b^cd^ 

20  ^-^^?^=  22.  12am^ri*^^^ 

Ixy^!^  ^rri?r^p^(f 

23.  (4a2^^z«Xl0a2i3  2)^5^3  52^^ 

24.  (21a,-2y*2«-3V^)(-2^y'z)- 

25.  104a5»a;»--(91a^^^^a;^H-7a^^.*:r)  = 

26.  (24an«:r--3a2^2)^(^35^6^2^^_5^3j^)^ 

27.  85a^"'  +  ^-i-5a*"*-2=  28.    84 a'*-* -4- 12^2 


48  ALGEBRA. 


Op  Polynomials  by  Monomials. 

95.  Tlie  product  of  (a  +  S  +  ^)  X  p  =  ap  -\-hp  -\-  cp. 

If  tlie  product  of  two  factors  be  divided  by  one  of  the 
factors,  tbe  quotient  is  the  other  factor.     Therefore, 

{ap  -\-hp-\-  cp)  -i-p  =  a-\-h-{-c. 

But  a,  h,  and  c  are  the  quotients  obtained  by  dividing 
each  term,  ap,  hp,  and  cp,  by  p. 

Therefore,  to  divide  a  polynomial  by  a  monomial, 

96.  Divide  each  term  of  the  polynomial  hy  the  monom^ial. 


Exercise  XXI. 

L  (8a5  — 12ac)-^4a  =  2i  — 3c. 

2.  (15am  — 10Sm  +  20cm)~  — 5m  =  — 3a  +  2Z>  — 4tf. 

3.  (ISamy— 27Z>7iy  +  36cpy)-f-  — 93/  = 

4.  {^\ax  —  \^lx-^\^cx)-^  —  Zx  = 

5.  iyis?--^o?^\x)---^x  = 

6.  (3a;«-6:^  +  92:^-12^^)-f-3a;2^ 

7.  (35m>  +  28mV-14w3/2)-^--7my  = 
a  (4a^5-6aS52+12a25«)-^2a25  = 

9.  (12a;«3/3-15a;V-24^3/)---3a;2^  = 

10.  (12a;*2/*-24a:V  +  36a;32/3-12^y2)--12a;2^  = 

11.  (3a*-2a'^5-a«^2)^^4_ 

12.  (3a;3y22  _|_  ^^4^^  _  15^3/2/ 4.  ISarV^)  ---  -  Sa:^^^  ^ 

13.  (-16a3^»V  +  8a^Z*2  6'^-12a^^>V)-f--4a2^'V  = 


DIVISION.  49 


Of  Polynomials  by  Polynomials. 

97.    If  the  divisor  (one  factor)  =  a  +  5  -f  e, 

and  the  quotient  (other  factor)  =  n  +^9  +  q, 

an-\-hn-{-cn 
then  the  dividend  (product)     =1  -{- ap -\- bp '\- cp 


+  aq-\-hq-{-  cq. 


The  first  term  of  the  dividend  is  an;  that  is,  the  product 
of  a,  the  first  term  of  the  divisor,  by  n,  the  first  term  of  the 
quotient.  The  first  term  n  of  the  quotient  is  therefore 
found  by  dividing  an,  the  first  term  of  the  dividend,  by  a, 
the  first  term  of  the  divisor. 

If  the  partial  product  formed  by  multiplying  the  entire 
divisor  by  n  be  subtracted  from  the  dividend,  the  first  term 
of  the  remainder  ap  is  the  product  of  a,  the  first  term  of 
the  divisor,  by  p,  the  second  term  of  the  quotient.  That  is, 
the  second  term  of  the  quotient  is  obtained  by  dividing  the 
first  term  of  the  remainder  by  the  first  term  of  the  divisor. 
In  like  manner,  the  third  term  of  the  quotient  is  obtained 
by  dividing  the  first  term  of  the  new  remainder  by  the  first 
term  of  the  divisor,  and  so  on. 

Therefore,  to  divide  one  polynomial  by  another, 

98.  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor. 

Write  the  result  as  the  first  term-  of  the  qtcotient. 

Multiply  all  the  terms  of  the  divisor  by  the  first  term  of 
the  quotient. 

8ubtraet  the  product  from  the  dividend. 

If  thei'e  be  a  remainder^  consider  it  as  a  new  dividend 
and  proceed  as  before. 


50  ALGEBRA. 


99.  It  is  of  great  importance  to  arrange  both  dividend 
and  divisor  according  to  the  ascending  or  descending  pow- 
ers of  some  common  letter^  and  to  Iceep  this  order  throughout 
the  operation. 

EXEECISE  XXII. 
Divide 

(1) 


a'-\-2ab-\-b^\)j  a-\-b]          (2) 

a'~Phj  a  +  b 

a^  +  2ab-\-b''\a-\-b 
o?-\-    ab          a-\-b 

ab  +  b^ 

ab  +  b^ 

a'-    P\a  +  b 

c^-\-ab  a  —  b 

-ab-y 

-ab-b^ 

(3)  c?—2ab-\-b'^hji 

.-5; 

a^-2ab-\-b^\a 

-b 

c?—    ah           a 

-  abJ^V 

-  ab-VW 

-b 

(4)  ^a^x" -Aa^x*  J^x^-a^hyi 

r^-a-; 

x^-4:a'x'  +  4:a''x'-a'\x^- 
x^—    o?x^                         x^- 
-Sa^x'  +  ^a'a^-a^ 
-Sa'x'  +  ^a'x' 

a'x'-a^ 
a'x'-a' 

-    a^ 

(5)  22aH''-\-lbb^  +  ?>a^-10a^b-22ab^hYd'-^W-2ab 

Sa'  -  lOa^b  +  22a'b^  -  22ab^+15b'\   a'-2ab-i-Sb' 
3a'-    6a^b+    9a'b' 3a'-4:ab  +  5b' 

-  4:a^b  +  13a'b'-22ab^ 

-  4:a^b+    Sa^b^-12ab^ 


ba^b^-10ab^+15b* 
baH^-10ab^+16b^ 


DIVISION.  51 


Divide 

6.  x'-lx-^Uhjx-^,. 

7.  a^-{-x-12hjx  +  ^. 

a  2x'-ct?-\-2>x-^hY2x-2>. 

9.  Ga,^  +  14x2_4^_|.24by2:r  +  6. 

10.  2>x^-{-x-^S)o^-lhjZx-l. 

11.  17^  +  b^x-2^x'-2\hY1x-^, 

12.  :r^— 1  byo;—  1. 

13.  a3-2aZ;2^i3by^__5^ 

14.  x'^  —  ^lif  hj x  —  ^y. 
1^.  x^  —  t/hY  x  —  y. 

16.  a''  +  32i*bya  +  2J. 

17.  2a^  +  27aZ>3-81^>^bya  +  36. 

18.  a^*  +  ll:r2-12a;-5a;3  +  6by3  +  :?,'2_3^^ 

19.  x^-S)x^-{-a?-lQ>x-4.hYX'-\-^-\-^x. 

20.  3G4-a:*-13:r'by6  +  a.-2+5x 

21.  a;'*  +  G4byr^  +  4:i-  +  8. 

22.  a;*  +  a;3  +  57-35.r-24a;2by.^2-3  +  2a:. 

23.  l-a;-3^-2-a;''byl  +  2:r  +  a.-2. 

24.  x^-2:^-{-\hYC(^-2x  +  l. 

25.  a4  +  2a252  +  9Z»^bya2-2a^>  +  3^'l 

26.  4^^  +  4:^-3  by  2  +  22-2^3,,_ 

27.  a' -243  by  a -3. 

28.  18^;*  +  82 :r  +  40  -  ^Ix-^ba^  by  3a,-2  +  5  -  4:r. 

29.  a;-*- G.ry- 9:^-2 -y2  by  :i-2  +  y  +  3:r. 


52  ALGEBRA. 


30.  a;^  +  9a:y-6rrV-4yny:r2-3;r?/  +  23/2^ 

31.  x^ -\- oc^ 'if' ■\- y"^  \iY  x^  —  xy  ■\- y^ . 

32.  x*  -\-7?-\-x^y-\-'if'  —  2xy^  —  oi?'if\)j  ci?^x  —  y. 

33.  'lx^-Zf-\-xy  —  xz  —  ^yz  —  '^\ij'2.x^Zy-\-z. 

34.  12  +  82:r2  +  106:r^-70a;^-112a^-38^ 

by  3  —  5:^4-  7:rl 

35.  3^-\-f\)jx'^  —  :^y^:i^'if~xf-\-y^. 

36.  2a:*  +  2:r2y2_2a;2/^-7:c3y-yny2a;2_|_^2_^^^ 

37.  16a:*  +  4a:2^2_|_^4i3y4^_2^y_|.y2^ 

38.  32a^^4-8a«53-a^'^-4a2  5^-56a^Z>2 

39.  l  +  5a:3_e^4]^y;^_^_^3^^ 

40.  l-52a^5^-51a^Z>3by4a252  +  3a5-l. 

41.  x*  y  —  xy''  hj  0(^y  -\-2xi^  —  2o(?'i^  —  y^. 

42.  a;«+15a;V+15:r2y*  +  /-6:c^y-6:ry^-20:rV 

bya:3_3^y_|_3^^2_^^ 

43.  a^  +  2a«^'^-2a*^3-2a«S-6a2  5^-3a&« 

bya3__2a2^>-G5l 

44.  ^lx^y  +  ld>x'f-b4:a^f-l^o(^y^-\^xf-^y' 

hj^x'-{-3?f-\-y\ 

45.  a*  +  2a3 ^  +  Sa^J^  +  l^ab^  +  165^  by  a^  +  452^ 

46.  8y«-2'«  +  21a;^2/3_24^^by3^^_^_^, 

47.  16a^  +  95^  +  8a252by4a2  +  3Z;2_4a6. 

48.  a^-\-h^  +  (?-^ahchja-\-h^c. 

49.  «»+  853+V-  6a&c  by  o?+W  +  c-^-  etc-  2a6  -2.be. 

50.  a^  +  ^'  +  c^+3a2Z>  +  3a^>2i3^^^^_l_^^ 


DIVISION.  53 


100.    The  operation  of  division  may  be  shortened  in  some 
cases  by  the  use  of  parentheses.     Thus : 

x^-i-(a-}-h-\-c)3^-\- {ah -{-ac-\-hc)x-\- abc \x  -{-h 


3?+{   -\-h       )x' a^-{-{a-\-c)x-{-ac 

(a        -{-c)oi^-\-{ab-{-ac-\-hc)x 
(g        -\'C)3(^-\-{ah  -\-hc)x 

acx  +  «^^ 

acx  -\-  dbc 


Exercise  XXIII. 
Divide 

1.  a^  {h  +  c) -]-W  {a-  c)  ^  (? {a-h)  +  ahchy  a'\-h -{- c. 

2.  a?  —  {a -{- h -\- c)  a? -{- {ah '\- ac -\- be)  X  —  dbc 

hj  a^  —  {a -{- h)  X -{•  ah. 

3.  a^  -2ax^  +  {a^  -\-  ah  -h^  X-  a^b  -\-  ah^  hj  X  -  a^  b. 

4.  x^  —  {a?  —  h  —  c)  a^  —  {b  —  c)  ax  -\-  he  hj  a^  —  a^  -{-  c. 

5.  y^—  {m-[-n-\-p)'ir-[-{mn-\-7np-{-np)y—innp\>jy—p. 

6.  x^  -{-  {b-^  a)  a^  -  {^-ba  +  h)  a?  -  {^a  +  bb)  x  +  4:h 

hy  a^ -{- b  X  —  ^. 

7.  x*-{a+h-{-c-{-  d)a?-^{ab  +  ac+ad-{-he-\-hd-\-ed)a^ 

—  {abc-\-ahd-\-acd-[-hcd)x-{-ahcd 

by  a?—{a-\-e)x-{-ae. 

8.  x^  —  {m  —  c) x^ -[- {n  —  em-]- d) a? -\- {r -{- en—  dm) a? 

■\- {er  •]■  drC)  X -{■  dr  ^yy  a?  —  ma?-\-nx-\'r, 

9.  .^•*  —  mx^  +  no^  —  na?  +  tnx  —  1  by  a;  —  1. 

10.    {x-{-yy-{-^{:x-\-yfz  +  ^{x-\-y)^  +  ^ 

^7{^  +  yY  +  ^{x  +  y)z^z\ 


54  ALGEBRA. 


101.  There  are  some  cases  in  Division  wliicli  occur  so 
often  in  algebraic  operations  that  they  should  be  carefully 
noticed  and  remembered. 


Case  I. 

The  student 

may  easily  verify  the  following 

results : 

(1) 

a  — 

■  +  ab  +  b\   ■ 

(2) 

27a} 

o  „. 

O  7, 

=  da^  +  6ah  +  4:P. 

(3)   ^L^  =  a'  +  a'b  +  a'h'  +  ab'  +  h\ 
(4) 


a  — 2b 


From  these  results  it  may  be  assumed  that : 

102.  The  difference  of  two  equal  odd  powers  of  any  two 
numbers  is  divisible  by  the  difference  of  the  numbers. 

It  will  also  be  seen  that : 

I.  The  .number  of  terms  in  the  quotient  is  equal  to  the 
exponent  of  the  powers. 

II.  The  signs  of  the  quotient  are  all  positive. 

III.  The  first  term  of  the  quotient  is  obtained,  as  usual, 
by  dividing  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor. 

IV.  Each  succeeding  term  of  the  quotient  may  be  ob- 
tained by  multiplying  the  preceding  term  of  the  quotient 
by  "^  second  term  of  the  divisor  (disregarding  the  sign), 
and  dividing  the  product  by  the  first  term  of  the  divisor. 


DIVISION.  65 


Exercise  XXIY. 
Write  by  inspection  the  results  in  the  following  examples 

1.  (f-l)^{y-l).  5.    {:^-y^)^{x-y). 

2.  {¥-12b)-^{h-b).  6.    (a*  -  1) -^  (a  -  1). 
(a -6).             7.    (l-8a;3)--(l-2a;). 
(.r-7).  8.    {:x^-2>2h')-^{x-2h). 

9.    {^o?x^-l)^{2ax-l). 

10.  {l-21x'f)^{l-2>xy). 

11.  (64  a« 53  _  27  ^)  -f-  (4  a6  -  3  rr). 

12.  (243  a* -1)-- (3  a -1). 

13.  (32a*-2436'')^(2a-3^). 


3.  (a3-216) 

4.  "(oi^  -  343) 


Case  II. 

(1)   <f±^  =  a'-uh^b\ 
a  +  o  . 

27^Hi8^^9^_6      _^4yi^ 
^  ^     3a;  +2y  ^  '     ^ 

a  +  o 

(4)    '?!^^^±^^  =  ^lx'-b^^y^2>Q>x'f-2^xf-\-lQy\ 
3^  +  2?/ 

From  these  results  it  may  be  assumed  that :    . 

103.  The  sum  of  two  equal  odd  powers  of  two  numbers  is 
divisible  by  the  sum  of  the  numbers. 

The  quotient  may  be  found  as  in  Case  I.,  but  the  signs 
are  alternately  plus  and  minus. 


66  ALGEBRA. 


Exercise  XXV. 
Write  by  inspection  the  results  in  the  following  examples : 

1.  Q^  +  f)-^{x-\-y).  5.    {^aH^-\-l)^{2ax-\-l). 

2.  (^•^  +  y^)-^(x  +  y).  6.    (a;«  +  272/«)-(a;  +  3y). 

3.  (l  +  8a^)--(l  +  2a).        7.    {c^ +  mh')  ~{a-\-2h). 

4.  {21a^  +  b^)~{Za-\-h).     8.    {bl2a?f-^z^)~-{^xy  +  z). 

9.    (729a3  +  216^3)--(9a  +  6Z>). 

10.  (64a3+1000Z>3)--(4a  +  10^). 

11.  (64a3  53  +  272:«)--(4aZ»  +  3a.-). 

12.  (a;3  +  343)H-(:r+7). 

13.  (27a;«y3  +  8  2^)  -^  (3  ^y  +  2  z), 

14.  (1024  a^  +  243  ^^^)  --  (4  a  +  3  h). 


Case  III. 
^-f^.^o,  m   "tut. 


(1)    -^ ^^x  +  y.  (2)    ^^^ ^-  =  a?  +  a^y  +  x7^-^if. 

x—y  x—y 

(3)    ^IlX'  =  :r-2/.  (4)    ^'^^^-r^y  +  ^y^-^/^. 

From  these  results  it  may  be  assumed  that : 

104.  The  difference  of  two  equal  even  powers  of  two  num- 
bers is  divisible  by  the  difference  and  also  by  the  sum  of  the 
numbers. 

When  the  divisor  is  the  difference  of  the  numbers,  the 
quotient  is  found  as  in  Case  I. 

When  the  divisor  is  the  sum  of  the  numbers,  the  quotient 
is  found  as  in  Case  II. 


DIVISION.  57 


Exercise  XXVI. 
Write  by  inspection  the  results  in  the  following  examples : 

1.  {x'-y')~{x~y).  a    (16^^-l)-(2:r-M). 

2.  {x'-y^)^{x  +  y).  9.    {^la'x'-l)-~{2>cix-\). 

3.  {a^-x^)^{a-x).  10.    (81a^a;*-l)--(3aa:+l). 

4.  (a^-x^)---{a  +  x).  11.   (64a«  -  5«) -- (2a- i). 

5.  {x'--^\y')-^{x~2>y).       12.    (64a«- Z/«)  -  (2a  + 5). 

6.  (:r^-81y*)--(a:  +  3y).       13.    (:r«- 729/) -- (a;- 3y). 

7.  (16:t-^-l)--(2a;-l).        14.    (:r«- 729/) -- (a;  + 3y). 

15.  (81a^-166-^)--(3a-2c). 

16.  (81a*-16c^)--(3a  +  2c). 

17.  (256  a*  -  10,000)  -  (4  a  -  10). 

18.  (256  a* -10,000)- (4  a +  10). 

19.  (625  a,-* -1)-- (5  a: -1). 

Case  IV. 

It  may  be  easily  verified  that : 

105.  The  sum  of  two  equal  even  poivers  of  two  numhe7's 
is  not  divisible  by  either  the  sum  or  the  difference  of  the 
numbers. 

But  when  the  exponent  of  each  of  the  two  equal  powers 
is  composed  of  an  odd  and  an  even  factor,  the  sum  of  the 
given  powers  is  divisible  by  the  sum  of  the  powers  expressed 
by  the  even  factor. 

Thus,  x^  +  /  is  not  divisible  by  x-\-y  or  by  x  —  y,  but 
is  divisible  by  ar*  +  /. 

The  quotient  mgty  be  found  as  in  Case  II. 


58  ALGEBRA. 


EXEHCISE   XXVII. 

Write  by  inspection  the  results  in  the  following  examples  : 

1.  (x'  +  f)  ~  (x'  +  f).         6.    (x^  +  1)  -  (x'  +  1). 

2.  (a«+l)--(a2+l).  7.    (Q4:X^-\-f)---(4:xr'  +  f). 

3.  (a}'  +  ^^')^(a^  +  f).      8.    (64  +  a^) -- (4  +  a^). 

4.  (5i«  +  1)  -  (^;2  +  1).         9.   (729  a'  +  b')  -V-  (9  a^  +  52>)_ 

5.  (a^  +  5^)--(a^  +  ^')-     10-   (729  c«  +  l)-- (9^  +  1). 

Note.  The  introduction  of  negative  numbers  requires  an  exten- 
sion of  the  meanings  of  some  terms  common  to  arithmetic  and 
algebra.  But  every  such  extension  of  meaning  must  be  consistent 
with  the  sense  previously  attached  to  the  term  and  with  general  laws 
already  established. 

Addition  in  algebra  does  not  necessarily  imply  augmentation,  as 
it  does  in  arithmetic.  Thus,  7  +  (—  5)  =  2.  The  word  sum,  however, 
is  used  to  denote  the  result. 

Such  a  result  is  called  the  algebraic  sum,  when  it  is  necessary 
to  distinguish  it  from  the  arithmetical  sum,  which  would  be  obtained 
by  adding  the  absolute  values  of  the  numbers. 

The  general  definition  of  Addition  is,  the  operation  of  uniting 
-two  or  more  numbers  in  a  single  expression  written  in  its  simplest 
form. 

The  general  definition  of  Subtraction  is,  the  operation  of  finding 
from  two  given  numbers,  called  minuend  and  subtrahend,  a  third 
number,  called  difference,  which  added  to  the  subtrahend  will  give 
the  minuend. 

The  general  definition  of  Multiplication  is,  the  operation  of  find- 
ing from  two  given  numbers,  called  multiplicand  and  multiplier,  a 
third  number,  called  product,  which  may  be  formed  from  the  mul- 
tiplicand as  the  multiplier  is  formed  from  unity. 

The  general  definition  of  Division  is,  the  operation  of  finding 
the  other  factor  when  the  product  of  two  factors  and  one  factor  are 
given. 


CHAPTER  V. 

Simple  Equations. 

106.  An  equation  is  a  statement  that  two  expressions  are 
equal.     Thus,  4  2: -12  =  8. 

107.  Every  equation  consists  of  two  parts,  called  the 
first  and  second  sides,  or  members,  of  the  equation. 

108.  An  identical  equation  is  one  in  which  the  two  sides 
arc  equal,  whatever  numbers  the  letters  stand  for.  Thus, 
(x-}-b){x-b)  =  x''-b^. 

109.  An  equation  of  condition  is  one  which  is  true  only 
when  the  letters  stand  for  particular  values.  Thus,  x-{-b 
=  8  is  true  only  when  x  =  S. 

110.  A  letter  to  which  a  particular  value  must  be  given 
in  order  that  the  statement  contained  in  an  equation  may 
be  true  is  called  an  unknown  quantity. 

111.  The  value  of  the  unknown  quantity  is  the  number 
which  substituted  for  it  will  satisfy/  the  equation,  and  is 
called  a  root  of  the  equation. 

112.  To  solve  an  equation  is  to  find  the  value  of  the 
unknown  quantity. 

113.  A  simple  equation  is  one  which  contains  only  the 
first  power  of  the  unknown  quantity,  and  is  also  called  an 
equation  of  the  first  degree. 


J 


60  ALGEBRA. 


114.  If  equal  changes  he  made  in  both  sides  of  an  equa- 
tion, the  results  will  he  equal.  §  43. 

(1)  To  find  the  value  oi  xm  x-\-h^=a. 

£c  +  6  =  a; 
Subtract  &  from  each  side,     x-^h—h  =  a—'b\ 
Cancel  +  6  —  &,  x=a—h. 

(2)  To  find  the  value  of  a;  in  a;  —  5  =  a. 

x  —  h  =  (2; 
Subtract  —  6  from  each  side,  x—h  +  &  =  a  +  J ; 
Cancel  —  &4-6,  x=-a-\-h. 

The  result  m  each  case  is  the  same  as  if  h  were  trans- 
posed to  the  other  side  of  the  equation  with  its  sign 
changed.     Therefore, 

115.  Any  term  m.ay  he  transposed  from  one  side  of  an 
equation  to  the  other  provided  its  sign  he  changed. 

For,  in  this  transposition,  the  same  number  is  subtracted 
from  each  side  of  the  equation. 

116.  The  signs  of  all  the  terms  on  each  side  of  an  equa- 
tion may  be  changed ;  for,  this  is  in  efiect  transposing  every 
term. 

117.  When  the  known  and  unknown  quantities  of  an 
equation  are  connected  by  the  signs  +  or  — ,  they  may  be 
separated  by  transposmg  the  known  quantities  to  one  side 
and  the  unknown  to  the  other. 

118.  Hence,  to  solve  an  equation  with  one  unknown 
quantity, 

Transpose  all  the  terms  involving  the  unknown  quantity 
to  the  left  side,  and  all  the  other  terms  to  the  right  side: 


SIMPLE   EQUATIONS.  61 


combine  the  like  terms,  and  divide  both  sides  by  the  coefficient 
of  the  unhnown  quantity. 

119.    To  verify  the  result,  substitute  the  value  of  the 
unknown  quantity  in  the  original  equation. 

Exercise  XXVIII. 
Find  the  value  of  x  in 

1.  5a:- 1  =  19.  8.  16a: -11  =  7a; +  70. 

2.  3a:  +  6=12.  9.  24a: - 49  =  19a:- 14. 

3.  24a:  =  7a: +  34.  10.  3a:  +  23  =  78- 2a:. 

4.  8a: -29  =  26 -3a:.  11.  26 -8a:  =  80- 14a:. 

5.  12 -5a:  =  19 -12a:.  12.  13  -  3a:  =  5a:-3. 

6.  3a:  +  6-2a:  =  7a:.  13.  3a:- 22  =  7a:  +  6. 

7.  5a: +  50  =  4a: +  56.  14.  8  +  4a:  =  12a:- 16. 

15.  5a: -(3^ -7)  =  4a: -(6a: -35). 

16.  6.r  -  2  (^  -  4a:)  +  3  (5a:  -  7)  =  10a:  -  (4  +  16a:  +  35). 

17.  9a:-3(5a:-6)  +  30  =  0. 

18.  a:  -  7  (4  a:  -  1 1)  =  14  (a:  -  5)  -  19  (8  -  a:)  -  61 . 

19.  (a: +7)  (a: -3)  =  (a: -5)  (a: -15). 

20.  (a:-8)(a:  +  12)  =  (a:  +  l)(a:-6). 

21.  (a:-2)(7-a:)  +  (a:-5)(a:  +  3)-2(a:-l)  +  12  =  0. 

22.  (2a: -  7)  (a:  +  5)  =  (9  ~  2a:)  (4 -a:) +  229. 

23.  14-a:-5(a:-3)(a:  +  2)  +  (5-a:)(4-5a:)=45a:-76. 

24.  (a.-  +  5)2-(4-a;)2  =  21a:. 

25.  5  (a; -2)H  7  (a; -3)2  =  (3 a; -7)  (4a: -19) +  42. 


62  ALGEBRA. 


Exercise  XXIX. 


PROBLEMS. 


1.  Find  a   number  such  that  when    12  is  added  to  its 

double  the  sum  shall  be  28. 

Let  X  =  the  number. 

Then  2x=  its  double, 

and        2  a;  +  12  =  double  the  number  increased  by  12. 
But  28  =  double  the  number  increased  by  12. 

.•.2a;4-12  =  28. 

2a;  =28 -12, 
2a;  =  16, 
a!=    8. 

2.  A  farmer  had  two  flocks  of  sheep,  each  containing  the 

same  number.  He  sold  21  sheep  from  one  flock  and 
70  from  the  other,  and  then  found  that  he  had  left  in 
one  flock  twice  as  many  as  in  the  other.  How  many 
had  he  in  each  ? 

Let  X  =  number  of  sheep  in  each  flock. 

Then  a;  —  21  =  number  of  sheep  left  in  one  flock, 
and        a;  —  70  =  number  of  sheep  left  in  the  other. 
.  • .  x-2l=2{x-  70), 
a  -  21  =  2a; -140. 
jB  -  2a;  =  -  140  +  21, 
-    a;  =  -119, 
X  =      119. 

3.  A  and  B  had  equal  sums  of  money  ;  B  gave  A  $5,  and 

then  3  times  A's  money  was  equal  to  11  times  B's 
money.    What  had  each  at  first  ? 

Let  X  =  number  of  dollars  each  had. 

Then       a;+  5  =  number  of  dollars  A  had  after  receiving  $5 
from  B, 
and  x  —  5  =  number  of  dollars  B  had  after  giving  A  ^5. 


SIMPLE   EQUATIONS.  63 

.•.3(a:+5)=ll(a;-5), 
3a;+15=lla;-55. 
3a;-lla;  =  -55-15. 
-8a;  =  -70, 

Therefore,  each  had  $8.75. 

4.  Find  a  number  whose  treble  exceeds  50  by  as  much  as 

its  double  falls  short  of  40. 
Let  X  =  the  number. 

Then  3  re  =  its  treble, 

and        3  a;  —  50  =  the  excess  of  its  treble  over  50 ; 
also,       40  —  2x  =  the  number  its  double  lacks  of  40. 
.•.3a;-50  =  40-2x; 
3a;  +  2a;  =40 +  50. 
5a;  =90, 
a;  =18. 

5.  What  two  numbers  are  those  whose  difference  is  14, 

and  whose  sum  is  48  ? 
Let  X  =  the  larger  number. 

Then  48  —  a;  =  the  smaller  number, 

and       (B  —  (48  —  a;)  =  the  difference  of  the  numbers. 
But  14  =  the  difference  of  the  numbers, 

.-.a;- (48 -a;)  =  14. 
a; -48  + a;  =14; 
2a;  =  ^; 
a;  =31. 

Therefore,  the  two  numbers  are  31  and  17. 

6.  To  the  double  of  a  certain  number  I  add  14,  and  obtain 

as  a  result  154.     What  is  the  number  ? 

7.  To  four  times  a  certain  number  I  add  16,  and  obtain  as 

a  result  188.     What  is  the  number  ? 

8.  By  adding  46  to  a  certain  number,  I  obtain  as  a  result 

a  number  three  times  as  large  as  the  original  number. 
Find  the  original  number. 


64  ALGEBRA. 


9.  One  number  is  three  times  as  large  as  another.  If  I 
take  the  smaller  from  16  and  the  greater  from  30, 
the  remainders  are  equal.     What  are  the  numbers  ? 

10.  Divide  the  number  92  into  four  parts,  such  that  the 

first  exceeds  the  second  by  10,  the  third  by  18,  and 
the  fourth  by  24. 

11.  The  sum  of  two  numbers  is  20 ;  and  if  three  times  the 

smaller  number  be  added  to  five  times  the  greater, 
the  sum  is  84.     What  are  the  numbers  ? 

12.  The  joint  ages  of  a  father  and  son  are  80  years.    If  the 

age  of  the  son  were  doubled,  he  would  be  10  years 
older  than  his  father.     What  is  the  age  of  each  ? 

13.  A  man  has  6  sons,  each  4  years  older  than  the  next 

younger.  The  eldest  is  three  times  as  old  as  the 
youngest.     What  is  the  age  of  each  ? 

14.  Add  $24  to  a  certain  sum  and  the  amount  will  be  as 

much  above  $80  as  thfiffsum  is  below  $80.  What  is 
the  sum?  >    +  ^'^'   t  •"       -     1,i:^ 

15.  Thirty  yards  of  cloth  and  40  yards  of  silk  together  cost 

$  330 ;  and  the  silk  cost  twice  as  much  a  yard  as  the 
cloth.     How  much  does^  each  cost  a  yard  ? 

16.  Find  the  number  whose  double  increased  by  24  exceeds 

80  by  as  much  as  the  number  itself  is  less  than  100. 

17.  The  sum  of  $500  is  divided  among  A,  B,  0,  and  D.    A 

and  B  have  together  $280,  A  and  0  $260,  and  A  and 
D  $220.     How  much  does  each  receive? 

18.  In  a  company  of  266  persons  composed  of  men,  women, 

and  children,  there  are  twice  as  many  men  as  women, 
and  twice  as  many  women  as  children.  How  many 
are  there  of  each  ? 


SIMPLE   EQUATIONS.  65 


19.  Find  two  numbers  differing  by  8,  such  that  four  times 

the  less  may  exceed  twice  the  greater  by  10. 

20.  A  is  58  years  older  than  B,  and  A's  age  is  as  much 

above  60  as  B's  age  is  below  50.  Find  the  age  of 
each. 

21.  A  man  leaves  his  property,  amounting  to  $  7500,  to  be 

divided  among  his  wife,  his  two  sons,  and  three  daugh- 
ters, as  follows :  a  son  is  to  have  twice  as  much  as  a 
daughter,  and  the  wife  $500  more  than  all  the  chil- 
dren together.     How  much  was  the  share  of  each  ? 

22.  A  vessel  containing  some  water  was  filled  by  pouring 

in  42  gallons,  and  there  was  then  in  the  vessel  seven 
times  as  much  as  at  first.  How  much  did  the  vessel 
hold? 

23.  A  has  $72  and  B  has  $52.     B  gives  A  a  certain  sum ; 

then  A  has  three  times  as  much  as  B.  How  much 
did  A  receive  from  B? 

24.  Divide  90  into  two  such  parts  that  four  times  one  part 

may  be  equal  to  five  times  the  other. 

25.  Divide  60  into  two  such  parts  that  one  part  exceeds 

the  other  by  2-i. 

26.  Divide  84  into  two  such  parts  that  one  part  may  be  less 

than  the  other  by  36. 

Note  I.  When  we  have  to  compare  the  ages  of  two  persons  at  a 
given  time,  and  also  a  number  of  years  after  or  before  the  given 
time,  we  must  remember  that  hoth  persons  will  be  so  many  years 
older  or  younger. 

Thus,  if  a;  represent  A's  age,  and  2x  B's  age,  at  the  present  time, 
A's  age  five  years  ago  will  be  represented  by  x  —  b\  and  B's  by 
2  a;  —  5.  A's  age  five  years  hence  will  be  represented  by  a:  -|-  5 ;  and 
B's  age  by  2ar  +  5. 


66  ALGEBRA. 


i 


27.  A  is  twice  as  old  as  B,  and  22  years  ago  lie  was  three 

times  as  old  as  B.     What  is  A's  age  ? 

28.  A  father  is  30  and  his  son  6  years  old.     In  how  many, 

years  will  the  father  be  just  twice  as  old  as  the  son  ? 

29.  A  is  twice  as  old  as  B,  and  20  years  since  he  was  three 

times  as  old.     What  is  B's  age  ? 

30.  A  is  three  times  as  old  as  B,  and  19  years  hence  he 

will  be  only  twice  as  old  as  B.  What  is  the  age 
of  each  ? 

31.  A  man  has  three  nephews ;  his  age  is  50,  and  the  joint 

ages  of  the  nephews  is  42.  How  long  will  it  be  be- 
fore the  joint  ages  of  the  nephews  will  be  equal  to 
that  of  the  uncle  ? 

Note  II.  In  problems  involving  quantities  of  the  same  kind 
expressed  in  different  units,  we  must  be  careful  to  reduce  all  the  quan- 
tities to  the  same  unit. 

Thus,  if  X  denote  a  number  of  inches,  all  the  quantities  of  the  same 
kind  involved  in  the  problem  must  be  reduced  to  inches. 

32.  A  sum  of  money  consists  of  dollars  and  twenty-iive-cent 

pieces,  and  amounts  to  $20.  The  number  of  coins  is 
50.     How  many  are  there  of  each  sort  ? 

33.  A  person  bought  30  pounds  of  sugar  of  two  different 

kinds,  and  paid  for  the  whole  $2.94.  The  better 
kind  cost  10  cents  a  pound  and  the  poorer  kind  7 
cents  a  pound.  How  many  pounds  were  there  of 
each  kind  ? 

34.  A  workman  was  hired  for  40  days,  at  §  1  for  every  day 

he  worked,  but  with  the  condition  that  for  every 
day  he  did  not  work  he  was  to  pay  45  cents  for  his 
board.  At  the  end  of  the  time  he  received  $22.60. 
How  many  days  did  he  work  ?  i 


SIMPLE   EQUATIONS.  67 

35.  A  wine  merchant  has  two  kinds  of  wine  ;  one  worth  50 

cents  a  quart,  and  the  other  75  cents  a  quart.  From 
these  he  wishes  to  make  a  mixture  of  100  gallons, 
worth  $2.40  a  gallon.  How  many  gallons  must  he 
take  of  each  kind, 

36.  A  gentleman  gave  some  children  10  cents  each,  and 

had  a  dollar  left.  He  found  that  he  would  have  re- 
quired one  dollar  more  to  enable  him  to  give  them 
15  cents  each.     How  many  children  were  there  ? 

37.  Two  casks  contain  equal  quantities  of  vinegar ;  from 

the  first  cask  34  quarts  are  drawn,  from  the  second, 
5o  &      2Gxgallons ;  the  quantity  remaining  in  one  vessel  is 
1        now  twice  that  in  the  other.     How  much  did  each 
cask  contain  at  first  ? 

38.  A  gentleman  hired  a  man  for  12  months,  at  the  wages 
^     of  $90  and  a  suit  of  clothes.    At  the  end  of  7  months 

the  man  quits  his  service  and  receives  $33.75  and 
the  suit  of  clothes.  What  was  the  price  of  the  suit 
of  clothes  ? 

39.  A  man  has  three  times  as  many  quarters  as  half-dollars, 

four  times  as  many  dimes  as  quarters,  and  twice  as 
many  half-dimes  as  dimes.  The  whole  sum  is  $  7.80.. 
How  many  coins  has  he  altogether  ? 

40.  A  person  paid  a  bill  of  $15.25  with  quarters  and  half- 

dollars,  and  gave  51  pieces  of  money  altogether. 
How  many  of  each  kind  were  there  ? 

41.  A  bill  of  £  100  pounds  was  paid  with  guineas  (21  shil- 

lings) and  half-crowns  (2i  shillings),  and  48  more 
half-crowns  than  guineas  were  used.  How  many  of 
each  were  paid  ? 


A/ 


CHAPTER  VI. 

Factoes. 

120.  In  multiplication  we  determine  the  'product  of  two 
given  factors ;  it  is  often  important  to  determine  the /actors 
of  a  given  product. 

121.  Case  I.  The  simplest  case  is  that  in  which  all  the 
terms  of  an  expression  have  one  common, factor.     Thus, 

J        (1)    o?^xy  =  x{x^y). 

(2)  6a^+4a2  +  8a  =  2a(3a2  +  2a  +  4). 

(3)  I8a«5-27a2Z>2  +  36a5  =  9a5(2a2-3a6  +  4). 

Exercise  XXX. 
Resolve  into  factors : 

1.  ba^-Vba.  4.   ^x^y-Vla?f-\-'^xf, 

2.  6^34- 18^2 _  12a.  5.   y^-m/  +  bf  +  cy. 

3.  ^^x^-2lx  +  l4:.  6.   Q>aW-2la'h^-\-21o^b\ 

7.  54:r23/«+I08a;y-243^y. 

8.  45a;yo-90^/-360a;y. 

9.  70»y-140ay  +  210a/. 
10.  32  aW  +  96  a^¥  -  128  a«6». 


FACTORS.  69 


122.    Case  II.    Frequently  the  terms  of   an   expression 
can  be  so  arranged  as  to  show  a  common  factor.     Thus, 

(1)  x^  +  ax  +  bx-i-ab  =  {x^-{-ax)  +  (bx-{-ab),       • 

=  X  (x  -\-  a)  -{-  b  (x  -{-  a), 
=:(x  +  b)(x  +  a). 

(2)  ac  —  ad—bc-{-bd=(ac  —  ad)~(bc  —  bd), 

=  a  (c  —  d)  ^  b  (c  —  d), 
=  {a-b){c-d). 


Exercise  XXXI. 
Resolve  into  factors : 

1.  x^  —  ax  —  bx-{-ab.  6.  dbx —  dby-\-pqx—pqy. 

2.  ab-\-ay  —  by  —  if.  7.  cdoi?  +  adxy  —  bcxy  —  abf. 

3.  bc-{-bx—  ex  —  0^.  8.  ahcy  —  V^dy  —  acdx  +  bd^x. 

4.  mx -{- mn -{- ax -\- an.  9.  ax  —  ay  —  bx-\-by. 

5.  cdoi?  —  cxy  +  dxy  —  f.  10.  (?c?2^  —  cyz  +  dyz  —  y^. 

123.  The  square  root  of  a  number  is  one  of  the  two  equal 
factors  of  that  number.  Thus,  the  square  root  of  25  is  5 ; 
for,  25  =  5  X  5. 

The  square  root  of  d^  is  o? ;  for,  a*  =  a^  X  o?. 

The  square  root  of  a^iV  is  abc ;  for,  o?b'^(?  —  abc  X  abc. 

In  general,  the  square  root  of  a  power  of  a  jiumber  is  ex- 
pressed by  writing  the  number  with  an  exponent  equal  to 
one-half  the  exponent  of  the  power. 

The  square  root  of  a  product  may  be  found  by  taking 
the  square  root  of  each  factor,  and  finding  the  product  of 
the  roots. 


70  ALGEBRA, 


The  square  root  of  a  positive  number  may  be  either  posi- 
tive or  negative ;  for, 

a^  =  a  X  a, 
or,  (J?  ^^  —  aY.—a\ 

but  throughout  this  chapter  only  the  positive  value  of  the 
square  root  will  be  taken. 

124.  Case  III.  From  §  73  it  is  seen  that  a  trinomial 
is  often  the  product  of  two  binomials.  Conversely,  a  trino- 
mial may,  in  certain  cases,  be  resolved  into  two  binomial 
factors.     Thus, 

To  find  the  factors  of 

The  first  term  of  each  binomial  factor  will  obviously  be  x. 

The  second  terms  of  the  two  binomial  factors  must  be  two 
numbers 

whose  'product  is  12, 
and  whose  sutyi  is  7. 

The  only  two  numbers  whose  product  is  12  and  whose 
sum  is  7  are  4  and  3. 

.-.  a;2_f_7^_f_12  =  (a;  +  4)(:r  +  3). 

Again,  to  find  the  factors  oi  oi?  -\-^xy  -\-^y^. 

The  first  term  of  each  binomial  factor  will  obviously  be  x. 

The  second  terms  of  the  two  binomial  factors  must  be  two 
numbers 

whose  product  is  6  y^, 
and  whose  sum  is        5y. 

The  only  two  numbers  whose  product  is  ^if  and  whose 
sum  is  5y  are  Zy  and  2?/.         ^ 

:.x'-\-hxy^^f  =  {x-\-Zy){x-\-2y). 


FACTORS.  71 


Exercise  XXXIL 
Find  the  factors  of : 

1.  rr2+lla;  +  24.  11.  :^  +  IS  ax  +  SQ  a\ 

2.  :^  +  ll:r  +  30.  12.  f +19p7/ -\- ^Sp\ 

3.  2/2_|_i7y4.60.  13.  2^-j-29qzi-100^. 

4.  z2_|.i32_|_i2.  14.  a^  +  ba'  +  G: 

5.  cc^-\-21x  +  U0.  15.  2^  +  42^  +  3. 

6.  3/2_|_35y_^300.  la  a2^,2^i3^_|_32^ 

7.  52  +  235  +  102.  17.  a;«/+7a;y  +  12.    ' 

8.  :r2_^3^_|.2.  la  210+102^  +  16.      ^ 

9.  a;2_|.7^_|.6.  19.  a'  +  9ab  +  20b\ 
10.  a2  +  9a5  +  852.  20.  x^  +  9a^  +  20. 

21.  aV+14a5^  +  3352.     24.   5V  +  18  a5c  +  65  a'*. 

22.  a^(f'  +  7acx-\-10a^.       25,   7^6^  +  23  rs2  +  90  2^. ' 

23.  x'ffz'  +  ldxyz  +  ^S.     26.   mV  +  207/^V^^  + 51pY. 

125.    CaseJV.   To  find  the  factors  of 

x'-dx  +  20. 

The  second  terms  of  the  two  binomial  factors  must  be  two 
numbers 

whose  product  is  20, 
and  whose  sum  is      —  9. 

The  only  two  numbers  whose  product  is  20  and  whose 
sum  is  —  9  arc  —  5  and  —  4. 

.'.  x" -dx  +  20  =  {x  -  6)(x  -  4:). 


72  ALGEBRA. 


Exercise  XXXIII. 
Eesolve  into  factors : 

1.  x^'-lx  +  lO.  13.  a^^cr"  -  IS  ahc  + 22. 

2.  :r2-29:r  +  190.  14.  x^-l5x  +  50. 

3.  a2_23a  +  132.  15.  a:^- 20  a;  +  100. 

4.  52-30^  +  200.  16.  aV-21a^  +  54. 

5.  ^-43z  +  460.  17.  aV-16a^»2;  +  3952. 

6.  a:2_7^_|_5^  18.  «V-24acz  + 14321 

7.  ri;*-4aV  +  3a^  19.  :i^-20^  +  91. 

8.  ^^^2-8^+12.  20.  .^2-23^7  +  120. 

9.  z^  -  572  +  56.  .  21.  2^  _  53  2  4_  36O. 

10.  /-7/+12.  22.  x^-(a-\-c)x  +  ac.     ■ 

11.  ^y-27:ry  +  26.  23.  f z^ -28  ahi/z  + 187 f/b\ 

12.  a'b^-Ua'b'  +  SO.  24.  c'd' -  SO  abed  +  221  a^b' 


126.    CaseV.   To  find  the  factors  of 

a^  +  2x  —  3. 

The  second  terms  of  the  two  binomial  factors  must  be 

two  numbers 

whose  product  is  —  3, 

and  whose  sum  is       +  2. 

The  only  two  numbers  whose  product  is  —  3  and  whose 
sum  is  +  2  are  +  3  and  —  1. 

.-.  a;2_|_  2^  _  3  =  (^ -f  3)  (;^  _  1). 


FACTORS.  73 


Exercise  XXXIV. 
Resolve  into  factors : 

1.  :r2^6a:-7.  8,  a'  +  25a-150. 

2.  rr2  +  5a:-84.  9.  h^  +  Sb^-4:. 

3.  /+7y— 60.  10.  Z>V  +  3^c-154. 

4.  2/2+i2y_45.  11.  6-i«+156-^-100. 

5.  z^+llz-12,  12.  6-2  4-17(7-390. 

6.  2^+132 -140.  13.  a2  4_«_i32. 

7.  a2+13a-300.  14.  0:^^^  +  9:^/2-22. 

127.   Case  VI.   To  find  the  factors  of 

The  second  terms  of  the  two  binomial  factors  must  be 

whose  product  is  —  66, 
and  whose  sum  is        —    5. 

The  only  two  numbers  whose  product  is  —  66  and  whose 
sum  is  —  5  are  —  11  and  +  6. 

.\x'-5x-Q6  =  {x~ll)(x-{-^). 

Exercise  XXXV. 
Resolve  into  factors : 

1.  x'-Sx-2^  6.  a'-lba-lOO. 

2.  2/^-72/ -18.  7.  c^'^-dc^-lO. 

3.  a^-9x-3Q.  8.  a.-^- 8:?;- 20. 

4.  z^-llz-QO.  9.  3/2_5^y__50a2 

5.  2^-132-14.  10.  a'b^-Sab-4:. 


74  ALGEBRA. 


11.  aV-3aa;-54.  14.    7/V-5yV-84. 

12.  c26/2-24cc?-180.         15.    a252_i(3^j_35^ 

13.  aV-a^c-2.  16.    a;2_(^_5)^_^5^ 

We  now  proceed  to  the  consideration  of  trinomials  which 
are  perfect  squares.  These  are  only  particular  forms  of 
Cases  III.  and  IV.,  but  from  their  importance  demand 
special  attention. 

128.   Case  y II.    To  find  the  factors  of 

^+18^  +  81. 

The  second  terms  of  the  two  binomial  factors  must  be 

two  numbers  •,  ^    ^  •    oi 

whose  product  is  81, 

and  whose  sum  is        18. 

The  only  two  numbers  whose  product  is  81  and  whose 
sum  is  18  are  9  and  9. 

.•.a^+18x  +  81  =  (x-\-9')(x  +  9)  =  (x  +  dy. 

Exercise  XXXVI. 
Eesolve  into  factors : 

1.  a^+12x  +  S6.  8.   3/4+16yV  +  64s^ 

2.  :r2  +  28:^+196.  9.   y«  +  24^+144. 

3.  :r2  +  345;  +  289.  10.   s^z' +162xz  +  6bGl. 

4.  z'  +  2z+l.  11.    4:a'+12ah^  +  9b\ 

5.  2/2  +  200?/ +  10,000.  12.    dxY-^S0x7/z  +  25z^. 

6.  2''+ 14^2  +  49.  13.    dx^-^l2x2/  +  4:y^. 

7.  a^  +  S6x7/-\-S24:y^.  14.   4 a^^;^  +  20 aVy  +  25 o^y . 


FACTORS.  75 


129.   Case  YIII.   To  find  the  factors  of 

The  second  terms  of  the  two  trinomials  must  be  two 

numbers 

whose  product  is  81, 

and  whose  sutu  is    —  18. 

The  only  two  numbers  whose  product  is  81  and  whose 
sum  is  —  18  are  —  9  and  —  9. 

.\a^-18x  +  Sl  =  (x-d)(x-d)  =  (x-dy. 

Exercise  XXXVII. 

1.  a2_8a+16.  10.  '^xY -20 x'y'z  + 25 i/z". 

2.  a2_30a-j-225.  11.  Ux'^-Sxy';^ -\-fz*. 

3.  a^-SSx  +  Sei.  12.  9a^b'c'-6aP(^d-{-b'(^d^ 

4.  a:2_4o^_|-400.  13.  lQ>a^ -8xy-\-x^i/*. 

5.  y2_ioOy  +  2500.  14.  a^x*-2(^ba^i/*-{-hy. 

6.  2/*-20?/2-f.lOO.  15.  S6a^f-mx7/  +  25y\ 

7.  9/-60i/z  +  (j252^.  16.  1 -6a53  +  9a25«. 

8.  x^ -32x^7/ +  2567/.  17.  97?zV-24mn  + 16. 

9.  2«-3423  +  289.  18.  4 ^» V -  12 J^c^y -f  9 a:^^. 

19.  49a2-112a6  +  64i2. 

20.  64a:y-160r5yz  +  100a:V. 

21.  ^9a'b^c'~28ahcx-i-4:x'. 

22.  121.r*-286;r2y+169y2. 

23.  289a:2/22_io2a;y¥(7+9y2z2J\ 

24.  361  x'/z^  -  76  abcxyz  +  4  c^^Z,^. 


76  ALGEBRA. 


130.  Case  IX.  An  expression  in  the  form  of  two  squares, 
with  the  negative  sign  between  them,  is  the  product  of  two 
factors  which  may  be  determined  as  follows : 

Take  the  square  root  of  the  first  number,  and  the  square 
root  of  the  second  number. 

The  sum  of  these  roots  will  form  the  first  factor ; 

The  difference  of  these  roots  will  form  the  second  factor. 
Thus : 

(1)  o?-h''  =  {a  +  h){a-h). 

(2)  a'-ih-  cf  =  \a-^{h  -  c)\\a-(h  -  c)\, 

=  \a-\-h  —  c\  \a  —  b-{-c\. 

(3)  {a-hf-{c-dy=\{a~h)+{c-d)\  \{a-h)-{c-d)], 

=  \a—b-\-c—d\  \a—h—c-\-d\. 

131.  The  terms  of  an  expression  may  often  be  arranged 
so  as  to  form  two  squares  with  the  negative  sign  between 
them,  and  the  expression  can  then  be  resolved  into  factors. 

Thus: 

a^J^b^-(^-d''-\-2ah-\-2cd, 

=^a^-\-2ah-\-Jy'-(^  +  2cd-d\ 

=  {p?  +  2ab  +  y)-{(P'  -2cd-\-  d% 

=  {a  +  hf-{c-df, 

=  \{a  +  h)  +  {c-d)]\{a-\-h)-{c-d)\, 

=  \a-\-h  +  c-d\\a  +  h-c  +  d\. 

132.  An  expression  may  often  be  resolved  into  three  or 
more  factors.     Thus  : 

(1)    x''-y''  =  {x^  +  y'){x'-f) 

=  {^  +  f)  {x'  +  y')  (x^  +  f)  (x'  -  f) 

=  {x^  +  y«)  ix'^  +  3/*)  {p?  ^if){x^  y)  {x  -  y). 


FACTORS.  77 


(2)   4(a5  +  c^)'-(a'  +  ^'-c''-(^7, 

\2(ab  +  cd)-(a'  +  P-c'-d')\, 
■=  \2ab  +  2cd-{-a'  +  b''-c'-d^ 

\2ab  +  2cd-  a^ -  P  +  c^ -{-  d^, 
=  \(a'  +  2ah  +  b^)-{c'  -  2cd+  d^)\ 

{{c"  +  2cd-^  d'')  -  (a? -  2ab  +  b^)\, 
=  \{a-\-bf-(^c-df\\{c  +  df-{a-bf\, 
=  \a  +  b  +  (c-d)\\a-{-b-(c-d)l 

lc-^d+(a-b)\\c  +  d-(a-b)i, 
=  \a  +  b  +  c-d\\a  +  b-c  +  d] 

\c  +  d+a-bl\c  +  d-a  +  bl. 

Exercise  XXXVIII. 
Resolve  into  factors : 

1.  a^-P.  14.  (ai-by-(c  +  dy. 

2.  a^-lG.  15.  {x  +  yy-(x-7jy. 

3.  4a2-25.  16.  2ab-a^-P+l. 

4.  a^-b\  17.  a^-2yz-i/-zK 

5.  a"-!.  18.  ar^-2x7/  +  7/-:^. 

6.  a^-b^  19.  a^+12bc-4:b^-9c^. 

7.  a»-l.  20.  a2-2ay  +  y2_^_o^2-22. 
a    36:^2-49^2.  21.  2a:?/-a;2_^_l_22 

9.  100 ar'f- 121  o'bK     22.  ^^-yS  _  ^_  ^2__2:ry-2c?2. 

10.  l-49ar^.  _23.  a;2_2^_|_22_  ^2  _2^2_|_2a2/. 

11.  a* -25^2  24.  2ab  +  a^~\-b^-(r'. 

12.  (a-5)2-c2.  25.  2:ry-ar^-y2  +  «2  +  ^,2_2a5. 

13.  a^-(a-by.  26.  (aa-  +  %)2-l. 


78  ALGEBRA. 


27. 

1- 

-^-f 

'  +  2X7J. 

31. 

{x 

4-1/ 

-{y- 

-If. 

28. 

(5 

a-'2.f- 

-(a -4)1 

32. 

cP 

—  3?  -^  ^xy  - 

-4/. 

29. 

0? 

-^ah  +  h^-x". 

33. 

a'- 

-¥- 

-2hc- 

-c'. 

30. 

{x 

+  1)^- 

(y+1/. 

34. 

Ax' 

'-^2 

.^4-6:^ 

;-l. 

133 

.   Case  X. 

Since 

0^- 
X- 

-4-^. 

^y  +  if. 

and 

a^- 

-!'--., 

A4- 

x'lF 

'  A-  xir 

^4-7/^ 

/ 

X  —  y  '^'      c/i^i^> 

and  so  on,  it  follows  that  tlie  difference  between  two  equal 
odd  powers  of  two  numbers  is  divisible  by  the  difference 
between  the  numbers. 


Exercise 

XXXIX. 

Besolve  into  factors : 

1.    a^-l\ 

6.   8a^-27y'. 

2.    a?-^. 

7.   647/2- 1000^3. 

3.   ^^-3-343. 

8.    729;r2-5127/2_ 

4.   7/2-125. 

9.  27^2-1728. 

5.   7/3  „  216. 

10.    1000  a^- 1331  Z»l 

134.    Case  XI. 

-  a?  —  ax -{- c^ , 

and                   ^-^y'- 

--x*~a 

:^y  4-  x^iP-  —  xifi  -\-  7/, 

x-^-y 

and  so  on,  it  follows  that  the  sum  of  two  equal  odd  powers 
of  two  numbers  is  divisible  by  the  sum  of  the  numbers. 


FACTORS. 


Exercise  XL. 
Resolve  into  factors : 

1.  7?-\-y'.  6.   216a3  +  512c8. 

2.  a?-^'^.  7.    729.r«4-17282/». 

3.  a;3  +  216.  8.   ar'  +  2/'. 

4.  2/^  +  042^.  9.   a;^  +  2/^ 

5.  64i='+125cf  10.   32^.^  +  2436^. 

) 
135.    Case  XII.  When  the  exponent  of  either  term  of 
an  expression  which  has  the  form  of  two  equal  even  powers 
with  the  plus  sign  between  them  can  be  resolved  into  arf^  q  ^a 
odd,  and  an  even  factor,  the  expression  is  divisible  by  the  J 
sum  of  the  powers  of  the  numbers  having  the  even  factor  j 
for  an  exponent.  ^^-<AC-i 

In  the  expression  .r*  +  if,  the  exponent  6  may  be  resolved 
into  two  factors,  3  and  2,  one  of  which  is  odd  and  the  other 
even,  and 

In  the  expression  x^  +  y*,  the  exponent  4  is  the  product 
of  two  even  factors,  2x2,  and  the  expression  cannot  be 
resolved  into  factors. 


Exercise  XLI. 
Resolve  into  factors : 

1.  a\-^h\         3.   o^^-\-y^.         6.   x''-^\.  7.    G4fi«  +  a;«. 

2.  ai«  +  ^i<>.       4.    6«  +  64c«.       6.    a^+1.         8.    729  +  c«. 


80  ALGEBRA. 


136.  Case  XIII.  For  a  trinomial  to  be  a  perfect  square, 
the  middle  tervi  must  he  twice  the  product  of  the  square 
roots  of  the  first  and  last  terms. 

The  expression  x"^  +  x^i/'  +  y^  will  become  a  perfect 
sqtiare  if  a^t^  be  added  to  the  middle  term.  And  if  the 
subtraction  of  x^y^  from  the  expression  thus  obtained  be  in- 
dicated, the  result  will  be  the  difference  of  two  squares. 
Thus, 

x""  +  x'y^  +  y^  =  {x^ -{-2x'y^  -{-  y^)  -  x'f, 
^{x^J^f)^-x?y\ 
=  {y?-\-f^xy){x^^y'^-xy), 
or,  {3? -^xy-^f){a?~xy-^  f)- 

Exercise  XLII. 
Eesolve  into  factors : 

1.  a^  +  «'^'  +  ^'-  8.   49m*+110mV  +  81w''.    ' 

2.  9^*  +  3:r2^2_^4y^  9.   9a^  +  21aV  + 25c^ 

3.  \<6x''-n:^f-\-y\  10.   A.'^a^ -V^a%'' ■\-Vl\h\ 

4.  81a*  +  23a2^2_^165^  11.   64^*+ 128^y  +  81?/^ 

5.  81a^-28a252+16^»4.  va.^^x'' -Zla^y'' ^^y\ 

6.  9a;*  +  38^?/2_|_49^4^  13^   '^^x" -^Xx'f -^\^y\ 

7.  25a^-9a2^2^16^>^  14.   ^Xx''  -?>^x'y'' -{-y". 

*  If,  in  Example  12,  9/  =  (-  Sy^f,  then  25 ar'/  should  be  added  to 
4a;*  —  Sls^y'^  +  93/*,  in  order  to  make  the  expression  a  perfect  square. 
That  is,  we  should  have  : 

(4  re*  -  12.^2/  +  9y*)  -  25  a;^^*, 
=  (2ar^- 32/2)2- 25  ar^/, 
=  (2ar2  -  3/  +  5a;,y)(2.r'  -  3/  _  5xy), 
or,  (2«2  +  5xy  -  3i/2)(2ar'  -  5 a^y  -  3;/). 


FACTORS.  81 


137.    Case  XIV.    To  find  the  factors  of 
6a^  +  x-l2. 

It  is  evident  that  the  first  terms  of  the  two  factors  might 
be  6:r  and  x,  or  2x  and  Sx,  since  the  product  of  either  of 
these  pairs  is  6a^. 

Likewise,  the  last  terms  of  the  two  factors  might  be  12 
and  1,  6  and  2,  or  4  and  3  (if  we  disregard  the  signs). 

From  these  it  is  necessary  to  select  such  as  will  produce 
the  middle  term  of  the  trinomial.  And  they  are  found  by 
trial  to  be  3  a:  and  2x,  and  —  4  and  +  3. 

.-.  Gx""  +  X -12  =  (Sx - 4:) (2x-i-S), 

Exercise  XLIII. 
Resolve  into  factors : 

1.  12a,-2-5a;-2.  13.  6aV  +  aa;-l. 

2.  12x2-7a;+l.  14.  Qb^'-lbx-Sx^. 

3.  12.^2  _^_  2^  15^  4^_|_8a7  +  3. 

4.  3x^-2x-6.  16.  a^-ax-~.6x^. 

5.  3a.-2  +  4.r-4.  17.  8a^-\-Uab-lbb^ 

6.  6x'i-5x-4:.  18.  Qa'-ldac  +  lOe'.    / 

7.  4:x'-{-lSx  +  S.  19.  83^  +  S4:X7/  +  21f. 

8.  Ax'^+llx-S.  20.  8ar'-22.ry-21y2. 

9.  4:a^-4:X-3.  21.  6ar^+19x7/-7f. 
16.   x'-Sax  +  2a\  22.  lla^- 23a&  +  2Z;2. 

11.  12a'  +  a-x^-x*.  23.   20"  ~13cd-\- 6d\ 

12.  2a,-2  +  5a.-y  +  2/.  24.    Qy'+lyz-S^. 


82  ALGEBRA. 


138.  Case  XV.  The  factors,  if  any  exist,  of  a  polyno- 
mial of  more  than  three  terms  can  often  be  found  by  the 
application  of  principles  already  explained.  Thus  it  is  seen 
at  a  glance  that  the  expression 

fulfils,  both  in  respect  to  exponents  and  coefficients,  the  laws 
stated  in  §  83  for  writing  the  power  of  a  binomial ;  and  it 
is  known  at  once  that 

a^-^a'b  +  2>ab''-W^{a-  bf. 

Again,  it  is  seen  that  the  expression 

a^  —  2x7/  +  3/^  +  2xz  —  2yz  +  z^ 

consists  of  three  squares  and  three  double  products,  and 
from  §  79,  is  the  square  of  a  trinomial  which  has  for  terms 
x,'y,  z. 

It  is  also  seen  from  the  double  product  —  2  a;y,  that  x  and 
y  have  unlike  signs  ; 

and  from  the  double  product  2xz,  that  x  and  z  have  like 
signs.     Hence, 

x^-2xy-\-f-\-2xz-2yz-\-^=^{x-y-\-zf. 

Exercise  XLIV. 
Kesolve  into  factors : 

1.  a^-\-2>a^b-\-2>ab^-{-b^.     4.   a;^+4r?;^y+6^y+4a7/+2/^ 

2.  a^  +  3a2  +  3a+l.  5.    ;r^- 4^;^  + 62:^- 4:r+ 1. 

3.  a^-3a2  +  3a-l.  6.    a^-4a«tf  +  6aV-4a6-3+6-''.3 

7.  x^-\-2xy-\-i/-{-2xz  +  2yz  +  ^. 

8.  x^-2xy  +  if-2xz-\-2yz-{-z'^. 

9.  a^  J^b^ ^ c^ ^2ab  ~2ac-2bc. 


FACTORS.  83 


139.  Case  XVI.   Multiply  2a:- y4- 3  by  a; +  2y- 3. 

2a:  -      ?/  +  3 
x+    2y-3 
2oi^—    xy  -\-Zx 

^xy-2f  +6y 

-6a:  +  3y-9 

2f*+3a:2/-2y*v-3a;  +  9y-9 

It  is  to  be  observed  that  2a*+3x3/  — 2?/',  of  the  product,  is  obtained 
from  {2x-y)  X  (x  +  2y) ; 

that  —  9  is  obtained  from  3  X  —  3  ; 

that  —  3  X  is  the  sum  of  2x  x  —  3  and  re  x  3 ; 

that  9y  is  the  sum  of  2?/  X  3  and  —  y  X  —  3. 

From  this  result  may  be  deduced  a  method  of  resolving 
into  its  factors  a  polynomial  which  is  composed  of  two  tri- 
nomial factors.     Thus : 

Find  the  factors  of 

^x" -Ixy -Zi/ -  9x-\-2>^y -21 . 
Tlio  factors  of  the  first  three  terms  are  (by  Case  XIV.) 
3a:  4-  y  and  2x  —  3y. 

Now  —  27  must  be  resolved  into  two  factors  such  that  the  sum  of 
the  products  obtained  by  multiplying  one  of  these  factors  into  3  a;  and 
the  other  into  2  a;  shall  be  —  9x. 

These  two  factors  evidently  are  —  9  and  +  3. 

That  is,  (<6a?  -  7a:y  -  S?/'  -  9a:  +  30y  -  27) 

=  (3a:  +  3/-9)(2a:-3y  +  3). 

140.  The  following  method  is  often  most  convenient  for 
separating  a  polynomial  into  its  factors : 

Find  the  factors  of 

2x'-bxy-^2f^1xz-bijz-\-2>:^. 

1.  Reject  the  terms  that  contain  z. 

2.  Reject  the  terms  that  contain  y. 

3.  Reject  the  terms  that  contain  x. 


84  ALGEBRA. 


Factor  the  expression  that  remains  in  each  case. 

1.  2a^-bxy  +  2y^=={x-2y){2x-y). 

2.  2x'^^1xz^-2>z^={x  +  ?>z){2x^z). 

3.  22/2-53/2  +  322  =  (_2y  +  32)(-2/+2). 

Arrange  these  three  pairs  of  factors  in  two  rows  of  three  factors 
each,  so  that  any  two  factors  of  each  row  may  have  a  common  term. 
Thus : 

x  —  2y,  x  +  2>z,  —2y^-?>z\ 

2x  —  y,  2x  +  Zj  —y+z. 

From  the  first  row,  select  the  terms  common  to  two  factors  for 
one  trinomial  factor : 

X  —  2y  +  3z. 

From  the  second  row,  select  the  terms  common  to  two  factors  for 
the  other  trinomial  factor : 

2x^y  ■\-  z. 

Then,  2a^-bx7/-\-2y^-\-7xz-57/z  +  3z^ 

=  (x~2y  +  Sz){2x-7/  +  z), 

141.  When  a  factor  obtained  from  the  first  three  terms 
is  also  a  factor  of  the  remaining  terms,  the  expression  is 
easily  resolved.     Thus : 

(3)   x^  —  Sxi/  +  2f-3x-i-67/, 
=.{x-2y){x-y)-Z{x-2yl 
=  (x-2y)ix-y-S). 

Exercise  XLV. 
Resolve  into  factors : 

1.  2x'-5xy  +  2y'-17xi-Uy  +  21. 

2.  6a^-S7xy  +  Qt/~5x-5y-l. 

3.  6x^  —  5xy  —  6'^  —  x  —  by  —  1. 

4.  6a^-Sxy  +  Sf+7x-5y  +  2. 

5.  2x^-xy-'Sy^-8x  +  1y  +  Q. 


FACTORS.  85 


6.  a;2  _  253/2  _i0a;-20y-f  21. 

7.  2oi?  —  bxy -{-^y^  —  xz  —  yz  —  '^. 

8.  ^x'-\-xy-f-Zxz-\-^yz-^^. 

9.  ^:x?~nxy^y^-\-Zbxz-byz-^^. 

10.  5ar^ -8a;y  + 32/2  _  3^2 +  3/2-22^. 

11.  'Ic^-xy-Zf-^yz  —  'l^. 

12.  ^3? -\Zxy -\-^y' -^Vlxz-lZyz^-^^. 

13.  a:2_2a:y  +  3/2_|_5^_5y^ 

14.  2:r2  +  5a:y-3y2_4a;2  +  22/z. 

Exercise  XL VI. 

MISCELLANEOUS   EXAMPLES. 

The  following  expressions  are  to  be  resolved  into  factors 
by  the.  principles  already  explained.  The  student  should 
first  carefully  remove  all  monomial  factors  from  the  ex- 
pressions. 

1.  5r^-15a;-20.  9.   a^  +  a^-fi. 

2.  2r^-16a;*  +  24A'«.  10.   o? -y" -xz-^yz.   t 

3.  3a2Z>2-9a5-12.  11.  ah- ac-h^^hc. 

4.  a^^'lax^o?-\-^a-\-^x:  12.    Zx? -Zxz-xy ^yz. 

5.  a2-2a5  +  ^'-^'  13.   c?-3?-ah-hx. 

6.  o?-2xy^i^-(?^'lcd-d'.  14.    a^ -'lax-Y3?-\-a-x. 

15.   3a;2_3^_2^4_2y/ 

a-^>.  16.   x''^o^-\-;i^-\-x. 

17.  aV-a2a;^-aV+l. 

18.  3:i.^-2:i;2y-27;ri/2+18?/3. 


86 


ALGEBEA. 


19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 
37. 
38. 
41. 
42. 
43. 
44. 
45. 
46 
47. 
48. 
49. 
58. 
59. 
61. 
62. 
63. 


729-/. 

a'*c  —  (^. 

a^  +  4:x~21. 

da^-21ah  +  30h\ 

2x^-ia^i/-6x'f.      36.    (a  +  bf-c^ 

s^—xi/—Qf-4:X-i-12y.    39.   3a;2_ix^^_|_g^^ 

1—x  +  a^-a^.  40.    a;2  +  20:r  +  91. 

(:r-2/)(.:2-22)-Gr-.)(a;2-y2). 

50.   2/2 -4y- 117.        / 


28.  4a2-4a&  +  Z^l 

29.  16a;2_gQ^^_|_]L002/2_ 

30.  36aVy2-255V^.     ^ 

31.  9.T2y*-302:y22j  +  2522. 

32.  16^  —  ^. 

33.  a;2-22:y-22:2;+y2+2y2+" 

34.  a2-a5-6Z^2_4^^i25. 

35.  .'r^  +  2rry  +  y2  — a;  — ?/— 6. 


a;2-5a:-24. 

6  xy +  ba^yz- 60  xz\ 

x^  —  2mx  -\-rn?  ~  r^. 

d!  +  a\ 

l-14A  +  49aV. 

o? -b^ - aiaj" -h^)  +  h  {a-hf. 

^3?f-Zx^-6y\  60.    6x'-^\?>xy-Y6f. 

6o?b^-ab^-l2b\ 

a!"  ^2ad-\-  d^  -  W  +\2bc  ~^(^. 

x^-2x'y-^^xif-%f.      64.   ^a^x^-^abx-Y^b\ 


51.  o?-\-6x-\Zh. 

52.  4a2_l2aS  +  9Z>'-4^. 

53.  {a-\-Uf-^{b-cf, 

54.  9x2_4^_|_4^2_^2_ 

55.  65V- 7 5:^3 -3a;^ 

56.  a'^-b^-Zabici-b). 

57.  a;^  +  2/^  +  3a:y(:r  +  y). 


FACTORS.  87 


65.  18.r2-24:ry  +  8y2  +  9:i;-6?/.  74.  16a^x-2x\ 

66.  2a^  +  2xi/~127/  +  6xz  +  18yz.  75.  S2ba^-4:bf. 

67.  {x  +  yy-l-xy(x  +  y+l),  76.  a;-27a;*. 

68.  :^-y'-z^  +  2i/z-\-x-\-^-z.  77.  o^-y^. 

69.  2:r2  +  4:ry  +  22/'  +  2aa:  +  2ay.  78.  4.9m^-l2ln\ 

70.  16  a2^,  +  32a^c  + 12^.6-2.  79.  16-81y*. 

71.  m^/>  —  m^^'  —  n^p  +  w^^^.  80.  12  2*  —  z^  —  6. 

72.  12aa:2_;L4^^y_g^^  81.  a;3_^_^^_i, 

73.  2a^  +  4:a^-10x.  82.   a;2_|_2a;  + 1  "2/^ 

83.  49(a-i)2-64(m-w)2. 

84.  i_A^-  +  ^>--A- 

V       2a6      J 

85.  :c2-53a;  +  360. 

86.  :^'3-2:r2y  +  a,•2-4:r  +  8y-4. 

87.  2ab-2bc  —  ae  +  ce  +  2b^-be. 

88.  125r'  +  350a;«y2+245:ry^ 

89.  a^  +  a'b-^-a^b^  +  a^b^+d'b^  +  ab'. 

90.  2a*a;  — 2a^ca;  +  2ac^a;  — 2c*a:. 

91.  6a;2-5a:y-6y2-f3a:2  +  15yz_9  22. 

92.  4:X^-9x7/  +  22/-Sxz-yz-z\ 

93.  3a2-7a5  +  2Z»2  +  5ac-56c  +  2c2. 

94.  x*-2x^  +  x^~8x  +  8. 

95.  5x2-8:r?/  +  3?/2_5a:  +  3y. 

96.  ^2 _ 2ac?+ c?2_ 4^2 _^  125c -9^2. 

97.  (x^-x-6)(x'-x-20). 


CHAPTER  VII. 

Common  Factors  and  Multiples. 

142.  A  common  factor  of  two  or  more  expressions  is  an 
expression  which  is  contained  in  each  of  them  without  a 
remainder.     Thus, 

5  a  is  a  common  factor  of  20  a  and  25  a  ; 
3:ry  is  a  common  factor  of  \2x^y'^  and  Iboi^i^. 

143.  Two  expressions  which  have  no  common  factor  ex- 
cept 1,  are  said  to  he  prime  to  each  other. 

144.  The  Highest  Common  Factor  of  two  or  more  expres- 
sions consists  of  all  the  factors  common  to  the  expressions. 

Thus,  3  a?  is  the  highest  common  factor  of  3  a^,  6  o?,  and 
12  a^ 

bxhf  is  the  highest  common  factor  of  lO^ry  and  Ibx^y^. 

For  brevity,  H.  C.  F.  will  be  used  for  Highest  Common 
Factor. 

(1)  Find  the  H.  C.  F.  of  42  a^b^x  and  21  a'bH^. 

^2a^b''x  =2xSx7xa^xPxx', 
21aWx^  =  ^Xl  X  a^  Xb^  X  x". 

:.  the  H.  0.  F.  =  3  X  7  X  a^  X  52  X  a;. 
=  2la%''x. 

(2)  Find  the  H.  C.  F.  of  2  A  -\-2ax^  and  3  abxy  +  3  bx'y. 

2a^x  -f  2aa^  =  2 ax  (a -\- x)  ; 
3  abxy  +  3  bx^y  =  3  bxy  (a  +  x). 

.'.  the  T[.G.¥.=x(a  +  x). 


COMMON    FACTORS   AND   MULTIPLES.  89 

(3)    Find  the  H.  C.  F.  of 

8  aV  -  24  A  +  16  a^  and  12  aa?y  — 12  axy  -  24  ay. 
8aV  _  ^^a^x  +  16^2  ^  g^a  ^^  _  3^ ^  2), 
=  2V(:i:-l)(a;-2); 
12ar2y  -  12aa;y  -  24 ay  =  Vlay  {3?-x  —  2), 

=  2^xSay(x+l)(x-2). 

.-.  theH.C.F.  =  22a(a:-2), 
=  4^(2; -2). 

Hence,  to  find  the  H.  C.  F.  of  two  or  more  expressions : 
Resolve  each  expression  into  its  lowest  factors. 
Select  from  these  the  lowest  power  of  each  corriTnon  factor, 
and  find  the  product  of  these  powei's. 

Exercise  XL VII. 
Find  the  H.  C.  F.  of : 
1.    l^aWc^d  and  ?>^a^bcd\        2.    I1p(f,  Up%  and  51/^. 

3.  8:cy2^  12a:y23,  and  20xyz'. 

4.  30  .^y,  90a;2^^  and  120  ary.  / 

5.  a'  -  h^  and  a^  -  h\  7.   a^  +  ^  and  [a  +  xf. 

6.  a^-x^Q.ndi{a-x)\  a    9:^- 1  and  (3a;+ 1)^. 

9.  13(?  -  4a;  and  la^x  -  ^a\ 

10.  12  aVy  -  4a3a:y2  and  30  a^a^f  -  10  ^2^3/3^ 

11.  8  a^b^c  -  12  a^^c^  and  6  ab^c  +  4  a5 V. 

12.  x^-2x  —  Ssindx^  +  x  —  12. 

13.  2a3-2a^2and45(a  +  5)2. 

14.  12:r33/(a;-2/)(a;-3y)andl8a:2(^^_^)(^3^_^) 

15.  ^.7^  +  6x^-24:X?Lnd6x^~96x. 


90  ALGEBRA. 


16.  ac  (a  —  h){a  —  c)  and  be  {h  ~  a)(b  —  c). 

17.  10:r«y  -  eOx'f  +  bxf  and  bx'f  -  bxf  -  lOO^/^ 

18.  x(x  +  iy,a^(a^-l)3^nd2x{x''-x-2). 

19.  30.-2-6^  +  3,  esr^  +  Qx-U,  and  VIt? -VI. 

20.  6  (a  -  5)^  8  (a^  -  ^2)2^  and  10  {a^  -  h% 

21.  x^  -  y^,  (x  +  yf,  and  a;^  _|_  3 ^^  _^  2y^ 

22.  a;^  —  y^,  x^  —  y^,  and  a^  —  7xy  +  63/^. 

■  23.  rt^-l,  a;3_i^  an(j^_^^_2.  ^ 

145.  When  it  is  required  to  find  the  H.  C.  F.  of  two  or 
more  expressions  which  cannot  readily  be  resolved  into 
their  factors,  the  method  to  be  employed  is  similar  to  that  of 
the  corresponding  case  in  arithmetic.  And  as  that  method 
consists  in  obtaining  pairs  of  continually  decreasing  numbers 
which  contain  as  a  factor  the  H.  0.  F.  required ;  so  in  alge- 
bra, pairs  of  expressions  of  continually  decreasing  degrees 
are  obtained,  which  contain  as  a  factor  the  H.  0.  F.  re- 
quired. 

The  method  depends  upon  two  principles : 

1.  Any  factor  of  an  expression  is  a  factor  also  of  any 
7)iultiple  of  that  expression. 

Thus,  if  F  represent  a  factor  of  an  expression  A,  so  that  A  =  nF, 
then  mA  =  innF.     That  is,  mA  contains  the  factor  F. 

2.  Any  common  factor  of  two  expressions  is  a  factor  of 
the  sum  or  difference  of  any  multiples  of  the  expressions. 

Thus,  if  F  represent  a  common  factor  of  the  expressions  A  and  B, 
so  that 

A  =  mF,  and  B  =  nF; 
then  pA  =  pmF,  and  qB  =  qnF. 

Hence,        pA  ±  qB  =pmF±  qnF, 
=  {pm  ±  qn)  F. 
That  is,  pA  ±  qB  contains  the  factor  F. 


COMMON    FACTORS   AND   MULTIPLES. 


91 


146.    The  general  proof  of  this   method  as  applied  to 
numbers  is  as  follows  : 

Let  a  and  b  be  two  numbers,  of  which  a  is  the  greater. 
The  operation  may  be  represented  by : 


/ 


b)a  (p 
pb 

-F)b(q 
qc 

d)  c  (r 
rd 


42)154(3 
126 

~28)42(l 
28 

14)28(2 
28 


nF)  mF  (p 
pnF 


~cF)nF(q 
qcF 

~F)cF(c 
cF 


p,  q,  and  r  represent  the  several  quotients, 
c  and  d  represent  the  remainders, 
and  d  is  supposed  to  be  contained  exactly  in  c. 
The  numbers  represented  are  all  integral. 
Then  c  =  rd, 

b^qc  +  d  =  qrd  +  d  =  {qr  +  1)  d, 
a  =pb  +  c  =pqrd  +pd  +  rd, 
=  {pqr  +p  +r)d. 

.'.  <f  is  a  common  factor  of  a  and  b. 

It  remains  to  show  that  d  is  the  highest  common  factor  of  a  and  b. 

Let/ represent  the  highest  common  factor  of  a  and  b. 

Now  c  =  a—pb,  and/  is  a  common  factor  of  a  and  b. 

.'.  by  (2) /is  a  factor  of  c. 

Also,  d=b  —  qc,  and / is  a  common  factor  of  b  and  c. 

.•.  by  (2)/  is  a  factor  of  d. 

That  is,  d  contains  the  highest  common  factor  of  a  and  b. 

But  it  has  been  shown  that  d  is  a  common  factor  of  a  and  b. 

.'.  d  is  the  highest  common  factor  of  a  and  b. 

Note.  The  second  operation  represents  the  application  of  the 
method  to  a  particular  case.  The  third  operation  is  intended  to  rep- 
resent clearly  that  every  remainder  in  the  course  of  the  operation 
contains  as  a  factor  the  H.  C.  F.  sought,  and  that  this  is  the  highest 
factor  common  to  that  remainder  and  the  preceding  divisor. 


92  ALGEBRA. 


147.    By  the  same  metliod,  find  the  H.  C.  F.  of 

2:r2+^-3  and  4:X^-i-8x^-x-6. 

'  2a^i-x-S)4.a^-i-8^-    ;r-6(2a;  +  3 
4^  +  2a;'-6^ 

6^  +  5a;-6 
6a^  +  dx~9 

2x  +  3)2x'+    x-S{x-l 
2x^  +  ?>x 
-2x-Z 
:.  the  H.  a  F.  -  2a;  +  3.  -  2.2;  -  3 


The  given  expressions  are  arranged  according  to  the  descending 
powers  of  x. 

The  expression  whose  first  term  is  of  the  lower  degree  is  taken 
for  the  divisor ;  and  each  division  is  continued  until  the  first  term  of 
the  remainder  is  of  lower  degree  than  that  of  the  divisor. 

148.  This  method  is  of  use  only  to  determine  the  com- 
pound factor  of  the  H.  C.  F.  Simple  factors  of  the  given 
expressions  must  first  be  separated  from  them,  and  the 
liighest  common  factor  of  these  must  be  reserved  to  be 
multiplied  into  the  compound  factor  obtained. 

Find  the  H.  C.  F.  of 

12a:*  +  30^-3  -  72n-2  and  32.^^  +  84:^-2 -  176a;. 

12a;*  +  30a.^-72a,-2=.6a;2(2^-j-5a;-12). 
Z2j?-^8^x^-VI^x-^/^x{8x'-\-2\x-^^). 
^0?  and  4  a;  have  2  a;  common. 

2a,'2  +  5a;- 12)8:^-2  +  21  a;-44(4 
8a.-2  +  20a;-48 

a;+   4)2a;2  +  5a;-12(2a;-3 
2a.-^  +  8a; 
-3a;-12 
.-.  the  H.  0.  F.  =  2x  {x  +  4).  -  3a;-12 


COMMON    FACTORS   AND   MULTIPLES.  93 

149.    Modifications  of  this  method  are  sometimes  needed. 
(1)    FindtheH.C.F.of4a;2_3^_5andl2ar^-4:r-65. 
4a;2_8^_5)12a;2_   4^_65(3 
120^-24^-15 
20  a; -50 

The  first  division  ends  here,  for  20a;  is  of  lower  degree  than  4  re*. 
But  if  20  a;  —  50  be  made  the  divisor,  4  a;*  will  not  contain  20  x  an  171- 
tegral  number  of  times. 

Now,  it  is  to  be  remembered  that  the  H.  C.  F.  sought  is  contained 
in  the  remainder  20  a;  —  50,  and  that  it  is  a  compound  factor.  Hence 
if  the  simple  factor  10  be.  removed,  the  H.  C.  F.  must  still  be  con- 
tained in  2  a;  —5,  and  therefore  the  process  may  be  continued  with 
2  a;  —  5  for  a  divisor. 


X- 

-5)4a;2_ 

-    8a:- 
"10a: 

-5( 

2a: 

+  1 

2a: 

-5 

2a'- 

-5 

,• 

.  the  H.  C.  F.  = 

2a:- 

-5. 

(2)    Find  the  H.  C.  F.  of 

21  a,'^  -  4ar^  -  15a:  -  2  and  2lx^  -  32ar'  -  54a:  -  7. 

21a;»-4a:2_i5^_2)21a:3-32a:2-54a:-7(l 
21a:«-    4a:^-15a:-2 
-28a-'2-39a:-5 

The  difficulty  here  cannot  be  obviated  by  removing  a  simple  factor 
from  the  remainder,  for  —  28 ar*  — 39a;—  5  has  no  simple  factor.  In 
this  case,  the  expression  21  ar'  -  4  a^  -  15  a;  —  2  must  be  midtiplied  by 
the  simple  factor  4  to  make  its  first  term  divisible  by  —  28  x^. 

The  introduction  of  such  a  factor  can  in  no  way  affect  the  H.  C.  F. 
sought ;  for  the  H.  C.  F.  contains  only  factors  common  to  the  remain- 
der and  the  last  divisor^  and  4  is  not  a  factor  of  the  remainder. 

The  signs  of  all  the  terms  of  the  remainder  may  be  changed ;  for 
if  an  expression  A  is  divisible  by  —  jP,  it  is  divisible  by  +  F. 


94 


ALGEBRA. 


The  process  then  is  continued  by  changing  the  signs  of  the  remain- 
der and  multiplying  the  divisor  by  4. 

28:u2^39:r  +  5)84r^-    IQar"-   60x-   8(3:r 
84:^^  +  117:^2+    15:r 


Multiply  by  —  4, 


Divide  by  -  63, 


-4 

•-   8 

532:1^  +  300  :^; 
532^:^^741^ 

+  32(19 

r+95 

-63) -441  a; 

-63 

Ix- 

7a;  + 1)280.-2  +  390:  + 5(4:^  +  5 
28:^2+   4a: 


■.  tlieH.aF.  =  7a:  +  l. 


35:r  +  5 
35.^  +  5 


(3)    Find  the  H.  C.  F.  of 

8;r2  _|- 2^  -  3  and  6:r3  +  5:^2  -  2. 
6^^+   5:^2-   2 


^  +  2a:-3)24a; 

^  +  20:^2_   3  (3^4_7 

24:r3+   6:^-   ^x 

14;r2+   ^x-   8 

Multiply  by  4, 

4 

56:r2  +  36a:-32 

560.^+14:^-21 

Divide  by  11, 

ll)22o:-ll 

2x-    l)8o:2_|_2^_3(4^-|-3 

8o:2_4^ 

6a:-3 

.-.  theH.C.F.: 

=  2o:-l.                               6o:-3 

In  this  case  it  is  necessary  to  multiply  by  4  the  given  expression 
Gar'  +  5a:2_2  to  make  its  first  term  divisible  by  8ar^,  4  being  obvi- 
ously not  a  common  factor. 


COMMON    FACTORS   AND    MULTIPLES. 


95 


The  following  arrangement  of  the  work  will  be  found 
most  convenient : 


Qx 
Qx 


4 


bx"-   2 


24:r«  + 20^:2  _    s 

9x~-    8 

56a;2_^36^_32 

56^:^+1437-21 

ll)22a;-ll 

2x-    1 


+  7 
4  07  +  3 


150.  From  the  foregoing  examples  it  will  be  seen  that,  in 
the  algebraic  process  of  finding  the  highest  common  factor, 
the  following  steps,  in  the  order  here  given,  must  be 
carefully  observed : 

I.  Simple  factors  of  the  given  expressions  are  to  be  re- 
moved from  them,  and  the  highest  common  factor  of  these 
is  to  be  reserved  as  a  factor  of  the  H.  0.  F.  sought. 

II.  The  resulting  compound  expressions  are .  to  be  ar- 
ranged according  to  the  descending  powers  of  a  common 
letter ;  and  that  expression  which  is  of  the  lower  degree  is 
to  be  taken  for  the  divisor;  or,  if  both  are  of  the  same 
degree,  that  whose  first  term  has  the  smaller  coefficient. 

III.  Each  division  is  to  be  continued  until  the  remainder 
is  of  lower  degree  than  the  divisor. 

IV.  If  the  final  remainder  of  any  division  is  found  to 
contain  a  factor  that  is  not  a  coynmon  factor  of  the  given 
expressions,  this  factor  is  to  be  rcTnoved;  and  the  resulting 

■expression  is  to  be  used  as  the  next  divisor. 

V.  A  dividend  whose  first  term  is  not  exactly  divisible 
by  the  first  term  of  the  divisor,  is  to  be  multiplied  by  such 
an  expression  as  will  make  it  thus  divisible. 


96  ALGEBRA. 


Exercise  XLVIII. 
FindtheH.C.F.  of: 

1.  5a;2  +  4a;-l,    20c(^  +  21x-5. 

2.  2:rS-4;r2-13a;-7,    Qx^ -lla^ -Six -20. 

3.  6a^  +  25a«-21a2  +  4a,    24a*+ 112a3- 94^^+ 18a. 

4.  da^  +  Qx'-Ax-A,    45:^  +  54^-20:r-24. 

5.  21x^-Sx^  +  6c(^-Sx',    162x^  +  4:8a^-18x^  +  6x. 

6.  20c(^-60x^  +  60x-20,    S2x^-92a^-\-68x'-24:x. 

7.  4:^-8.r-5,    12s^-4:X-65. 

8.  3a«-5A-2a:?;^    9  a^  -  8  a^a;  -  20  a^r^.       ^'' 

9.  10a^  +  x'-9x  +  24:,    20x* -17a^  +  i8x  ~S. 

10.  8:z^-4a;2-32:^-182,    S6c(^ -Six" -Ulx -126. 

11.  5a;2(12:r^+45^+17x-3),  10 ^(24 2:^-522:2 +14 a;-!). 

12.  9x^2/-x^7/^-20xy\    ISo^y -l8xY -,2xf -8y\ 

13.  6a;2-a7-15,    9:r2_3^_20. 

14.  l2a^-9o(^  +  bx  +  2,    2^:0^ ^lOx-\-l. 

15.  6ic3  +  15:z;2_g^_^9^    92,'^  + 6:^- 51^  + 36. 

16.  4:0^  —  oc^y  —  x^if  —  b'l^,    7:r^  +  4;r^y  +  4a:y^  — 3y^. 

17.  2a3-2a2-3a-2,    Sa^- a2_2a- 16. 

18.  12?/3_|_2y2_94y_60,    482/3 -24y2_  348?/ +  30. 

19.  9x{2x^-6x^~x^-^lbx-\(d), 

6a;2(4^4_|_e_^3_4^_15^_25)^ 

20.  15a;''  +  2a;3-75:r2+557+2,    35a;^+ri:3-175a;2+30:u  + 1. 

21.  21:u*-4a;3-15:i;2_2;^^    21  ^c^- 320.-2- 54 ;r- 7. 

22.  9x^y-22xh/-Zxy^-\-\0^f,   9^y-6a;y+.ry-25^-y«. 


COMMON   FACTORS   AND   MULTIPLES.  97 

23.  Qx'-4:X*-Ua^-Sx^-3x-l, 

24.  x^-aa^~a''a^-a^x-2a\    Sa^-7aa^-{-Sa^x-2a\ 

151.  The  H.  C.  F.  of  three  expressions  will  be  obtained 
by  finding  the  H.  C.  F.  of  two  of  them,  and  then  of  that  and 
the  third  expression. 

For,  HA,  B,  and  (7  are  three  expressions, 

and  D  the  highest  common  factor  of  A  and  B, 

and  -27  the  highest  common  factor  of  D  and  (7, 

Then  I)  contains  every  factor  common  to  A  and  B, 

and  U  contains  every  factor  common  to  I)  and  C. 
.'.  E  contains  every  factor  common  to  ^,  J5,  and  C. 

Exercise  XLIX. 
FindtheH.C.F.  of: 

1.  2x^-\-x-l,    x^^bx:\-^,    a?+l. 

2.  f-lf-y+l,    32/^-2y-l,    f-f^^y-l. 

3.  a^-^a^+'dx-lO,  c(^-^2x^-2>x-\-20,  a^^bx^-^x-^2>b. 

4.  x'-lx^  +  \Q>x-l2,    2>:f?-Ux'^l&x, 

5:^3 -10:^2  + 7a; -14. 

5.  y3_5y2+lly-15,    y3_2/2  +  3y^5^ 

6.  2x'-\-2>x-f),    Zo^-x-2,    2x?-\-x-Z. 

7.  a;3-l,    x^-x'-x-2,    23?-x'-x-2>. 

8.  7?-%x-2,    2x?-\-^x'-\,    x?^\. 

9.  \2{x^-y%f^{x^_-y''),    8(a;V  +  a:y5. 
\^,'x'-^xy\    x?y^y\    x'  +  x'f  +  y^ 

11.    2{y?y-xf),    Z{p?y-xf),    ^{x^y-xy%    5C^^y-a:/). 


98  ALGEBRA. 


Lowest  Common  Multiple. 

152.  A  common  multiple  of  two  or  more  expressions  is 
an  expression  which  is  exactly  divisible  by  each  of  them. 

153.  The  Lowest  Common  Multiple  of  two  or  more  ex- 
pressions consists  of  all  the  factors  of  the  expressions,  each 
factor  being  written  with  its  highest  exponent. 

154.  The  lowest  common  multiple  of  two  expressions 
which  have  no  common  factor  will  be  their  product. 

For  brevity  L.  0.  M.  will  be  used  for  Lowest  Common 
Multiple. 

^)    Find  the  L.  C.  M.  of  12  a^e,  Uhc^,  S6ab\ 
120^0  =  2^  xSa'c, 
Ube'  =  2  Xlbc^, 
2>^ah^^2^x2>''ab\ 

.-.  the  L.  C.  M.  =  22  X  32  X  Ta^^V  =  2b2a^b^(^. 

(2)    Find  the  L.  C.  M.  of 

2o?-\-2ax,^o?-^x^,^oj'-Q)ax-\-?>x^. 

2o?^2ax  =2a(a  +  :r), 

^a^-^x?  =2x3(a  +  :r)(a-.'r), 

3(2^  -  6aa:  +  3^2  =  3  (a  -  2:)l 

.'.  the  L.  C.  M.  =  6  a  (a  +  x)  (a  -  xj. 

Exercise  L. 
Find  the  L.  C.  M.  of : 

1.  ^a^x,    6ciV,    2ax'.  4.   :^-\    0? - x. 

2.  V^ax",    12ay\    I2xy.  5.    a?-b\    a?-^ab. 

3.  x",    ax-\-c^.  6.    2:r-l,    ^x^-l. 


COMMON   FACTORS   AND   MULTIPLES.  99 

7.  a  +  b,    o?-\-b\  9.    s^-x,   a?-l,   a?-\-l. 

8.  a^-l,   x'+l,   x"-!.        10.   a^-l,    o?-x,   a?-\, 

11.  2a +1,    4^2- 1,    Sa^  +  l. 

12.  (a +  5)2,    a^-h\ 

13.  4(1 +  a;),    4(1-0;),    "li^-o?). 

14.  a;-l,    a^  +  a;  +  l,    a;^-!. 

15.  x^-if,    {x^y)\    {x-y)\ 

16.  0?-^,    ^{x-y)\    12 (^r^ +  7/^). 

17.  G(:^+a:y),   K^y-f),    \^{^-f). 

18.  a,'2  +  5a:  +  6,    a;2_^6^_^8^ 

19.  a2-a-20,    a2  +  a-12. 

20.  a;2_|_ii^,_^30,    :r2  +  12a;  +  35. 

21.  a^-9a:-22,    .^2-13:^  + 22. 

22.  4aZ>(a2-3a5  +  252),    5a2(a2-j_a5  _  6Z>2), 

23.  20(a;2_i)^    24(a;2_^_2)^    l^ix'^x-l). 

24.  12;ry(:r2_yj)^    2^(.r  +  y)2,    3y2(:^_y)2. 

25.  {a  —  h)(h  —  c),    (b  —  c)(c  —  a),    {c  —  a)(a  —  h). 

26.  (a  —  b)(a  —  c),    {b  —  a){b  —  c),    (c~a)(c  —  b). 

27.  ar'-4ar^  +  3a;,    rr^  +  a,-^- 12^,    ^  +  3:c^-4a;3. 

28.  rr^y-V,    3a;(a;-y)2,    4:y{x-y)\ 

29.  (a+5)2-(c4-cf)«,   (a+c)2-(5  +  c^)2,   (a4-cZ)2_(54-c)2. 

30.  (2a;-4)(3:r-6),   (:r-3)(4a;-8),   (2x-Q)(5x-10). 

155.  When  the  expressions  cannot  be  readily  resolved 
into  their  factors,  the  expressions  may  be  resolved  by  find- 
ing their  H.  C.  F. 


100  ALGEBRA. 


Find  the  L.  0.  M.  of 


2>x'y-\-^xf^2f 
2>x^y  +  4.xf  +  2f 


^a?-22xi 

2 


V^3?-^^xy'-\^y' 
\%x?-^^x'y-\-    62/3 


ll?/)33a:^y-44a:?/-22: 


Zx?  -   ^xy  -  2y'^2x-y 

Hence,  6:^3-11  a;2y  + 2/=- (2ar-2/)  {^x'-^xy-^if), 
and  ^:x?-22xf-Sf^{^x  +  4.y){^x'-4.xy-2y^). 

.-.  the  L.  0.  M.  =- (2a; -y)(3:r +  4y)(3:f2- 4:ry  -  2y2). 

In  this  example  we  find  the  H.  C.  F.  of  the  given  expres- 
sion, and  divide  each  of  them  by  the  H.  C.  F. 

156.  It  will  be  observed  that  the  product  of  the  H.  0.  F. 
and  the  L.  C.  M.  of  two  expressions  is  equal  to  the  product 
of  the  given  expressions.     For, 

Let  A  and  B  denote  the  two  expressions,  and  D  their 
H.  C.  F. 

Suppose  A  =  aD,  and  B  =  hD  ; 

Since  D  consists  of  all  the  factors  common  to  A  and  B, , 
a  and  b  have  no  common  factor. 

.*.  L.  C.  M.  of  a  and  h  is  ah. 

Hence,  the  L.  0.  M.  of  aD  and  hD  is  ohD. 

Now,  A  =  aD,  d^nd  B=hD', 

.-.  AB  =  ahDxD. 

AB 

.'.  — =—  =  ahD  =  the  lowest  common  multiple.     That  is, 

The  L.  0.  M.  of  two  expressions  can  he  found  hy  dividing 
their  product  hy  their  H,  C.  F. 

Or,  hy  dividing  one  of  the  expressions  hy  the  H.  C.  F.,  cmc? 
onuliiplying  the  result  hy  the  other  expression. 


COMMON   FACTORS   AND   MULTIPLES.  101 

167.  To  find  the  L.  C.  M.  of  three  expressions,  A,  B,  C. 
Find  M,  the  L.  C.  M.  of  ^  and  ^ ;  then  the  L.  C.  M.  of  M 
and  C  is  the  L.  0.  M.  required. 

Exercise  LI. 
Find  the  L.  CM.  of: 

2.  a^-1,    x'  +  2x-S,    Qx'-x-2.  '^^''^^^  ^ 

3.  :i;3_27,    x'-l^x-i-SG,    a^-Sx'-2x  +  6. 

4.  5a.^+19a;-4,    10ar'-\-lSx-S. 

5.  Ux'  +  xij-Qif,    lSar^+18x^-20j/^. 

6.  x*-2x^  +  x,    2:?;*-2a:3_2a:_2. 

7.  120.-2  + 2a: -4,    12ar^-42a:-24,    12:z:2_  28a;_24. 

8.  a^-Qx'+llx-e,    a^-9ar'  +  26x--.24:,^ 

s^-8ar'-{-19x-12. 

9.  ar'-4a^  :>^+2ax'  +  4:a'x-{-8a^,  a^-2ax'+4a^x-Sa\ 

10.  a^  +  2a^i/-xy^-2y^,    a^-2x^i/ -xif +  2f. 

11.  l-hp+p',    l-p+p\    l+p'+p*. 

12.  (1-a),    (1-a)',    (!-«)'. 

13.  {a  +  cf-b^    (a  +  bf-c',    {hJ^cf-a\ 

14.  3c^-3c2y  +  cy2-2/3^    ^c^^c'y-Zcf. 

15.  m^-Sm  +  S",    m^  +  3m^  +  m  +  3.- 

16.  20n*  +  w2-l,    25w*+ 5^3-^-1. 

17.  ¥~2b^  +  ¥-8b  +  8,    U^-l2b^  +  9b-l. 

18.  2?'*-8r*+127^-8r2  +  2r,    2>7^ -Qt^ -\-2>r. 


CHAPTER    VIII. 

Fractions. 


158.  The  expression  -  is  employed  to  indicate  that  a 

units  are  divided  into  h  equal  parts,  and  that  one  of  these 
parts  is  taken ; 

or,  that  one  unit  is  divided  into  h  equal  parts,  and  that  a 
of  these  parts  are  taken. 

159.  The  expression  -  is  called  a  fraction,     a  is  the  nu- 

0 

merator,  and  h  the  denominator. 


160.  The  numerator  and  denominator  are  called  the 
terms  of  the  fraction. 

161.  The  denominator  shows  into  how  many  equal  parts 
the  unit  is  divided,  and  therefore  names  the  part ;  and  the 
numerator  shows  how  many  of  these  parts  are  taken. 

It  will  be  observed  that  a  letter  written  above  the  line 
in  a  fraction  serves  a  very  different  purpose  from  that  of  a 
letter  written  helow  the  line. 

A  letter  written  above  the  line  denotes  number ; 

A  letter  written  below  the  line  denotes  name. 

162.  Every  whole  number  may  be  written  in  the  form 
of  a  fraction  with  unity  for  its  denominator  ;  thus,  a  =  -. 


FRACTIONS.  103 


To  Reduce  a  Fraction  to  its  Lowest  Terms. 

163.  Let  the  line  AB  be  divided  into  5  equal  parts,  at 
the  points  C,  D,  E,  F. 

^1    I    I    I    I    I    I    I    I    I    I    I    I    I    I    I/? 
C  D  E  F 

Then^i^isfof^^.  (1) 

Now  let  each  of  the  parts  be  subdivided  into  3  equal 
parts. 

Then  AB  contains  15  of  these  subdivisions,  and  ^.F  con- 
tains 12  of  these  subdivisions. 

:.AFi^^oiAB.  (2) 

Comparing  (1)  and  (2),  it  is  evident  that  f  =  -J^. 

In  general : 

If  we  suppose  AB  to  be  divided  into  b  equal  parts,  and 
that  ^i^  contains  a  of  these  parts, 

Then^i^is^of^i?.  (3) 

o 

Now,  if  we  suppose  each  of  the  parts  to  be  subdivided 
into  c  equal  parts, 

Then  AB  contains  he  of  these  subdivisions,  and  AF  con- 
tains ac  of  these  subdivisions. 

.-.  ^i^isf^of^^.  (4) 

00 


Comparing  (3)  and  (4),  it  is  evident  that 


a  _ac 
b~  be 

Since  —  is  obtained  by  multiplying  by  c  both  terms  of 

the  fraction  -, 
b 

and,  conversely,  -  is  obtained  by  dividing  by  e  both  terms 

of  the  fraction  -— ,  it  follows  that 
be 


104  ALGEBRA. 


I.  If  the  numerator  and  denominator  of  a  fraction  be 
multiplied  by  the  same  number,  the  value  of  the  fraction 
is  not  altered. 

II.  If  the  numerator  and  denominator  be  divided  by  the 
same  number,  the  value  of  the  fraction  is  not  altered. 

r 

Hence,  to  reduce  a  fraction  to  lower  terms, 

Divide  the  numerator  and  denominator  by  any  common 
factor. 


164.  A  fraction  is  expressed  in  its  lowest  terms  when 
both  numerator  and  denominator  are  divided  by  their 
H.  0.  F. 

Reduce  the  following  fractions  to  their  lowest  terms  : 
-a^  __  (a  —  x) (a^ -\-ax-\-o(?)  __  a^-\-ax-{-0(^ 


(1) 


(2) 


(3) 


(4) 


0^  —  0?  {a  —  x^{a-\-x^  a-\-x 

a^  +  7a+10_(a  +  5)(a  +  2)^a  +  5 
a2  +  5a  +  6        (a  +  3)(a  +  2)       a  +  3* 

6r^-5a;-6  ^  (2:r  -  3)(3a;  +  2)  ^  3a;  +  2 
8:r2-2;r-15       (2;r-3)(4:i:  +  5)       4a;+5' 


a 


,3-7a2  +  16a-12 


3a3-14a2+16a 


Since  in  Ex.  (4)  no  common  factor  can  be  determined 
by  inspection,  it  is  necessary  to  find  the  H.  C.  F.  of  the 
numerator  and  denominator  by  the  method  of  division^ 

Suppress  the  factor  a  of  the  denominator"  and  proceed^^i^^^ide  : 

t 

i 


FRACTIONS. 

lUo 

a'-    1a'  +  16a-    12 

Sa'~ 

14a +  16 

a- 

-7 

3 

Sa'- 

6a 

3a«-21a2  +  48a-   36 

— 

8a+16 

3a- 

-8 

Sa^-UaJ'  +  lQa 

— 

8a+16 

-    7a2  +  32a-   36 

3 

-21a2  +  96a-108 

-21a2  +  98a-112 

-2)-   2a+     4 

a-     2 

.-.  the  H.  ( 

l¥.  =  a 

-2. 

Now,  if  a»  -  7a2  +  16  a  -  12  be  divided  by  a  -  2,  the  re- 
sult is  a^  —  5 a  +  i  ;  and  if  3a^  —  14a^  +  16a  be  divided  by 
a  —  2,  the  result  is  3  a^  —  8  a. 

a»-7a^  +  16a-12_a'-5a  +  6 


3a»-14a2+16a 


3a2-8a 


165.  "When  common  factors  cannot  be  determined  by  in- 
spection, the  H.  C.  F.  must  be  foujid  by  the  method  of 
division. 

EXEECISE   LIT. 

Reduce  to  lowest  terms : 


1. 


2. 


a^-1 


5. 


4.x  (x+l) 

a;^-9a;  +  20 

^^-7:^+12* 

a:^-2:r-3 
:r2-10a;  +  2l' 

x*  +  x^+l 
a^  +  x+l' 

+  2a:^?/  +  y^ 


6. 


7. 


a«+l 


10. 


a'^  +  2a2  +  2a 

+1' 

a2-a-20 

a2  +  a-12* 

a^-4.x'  +  9x 

-10 

x'  +  ^x'-Sx 

+  20' 

a^-bx^'+llx-l^ 

:i^-x''  +  3x 

+  5 

x*-\-a^^  +  xip 

-f 

2(?y  —  x\^  —  y* 


106  ALGEBRA. 


'    57^  +  ^  — ^  —  1' 


^-^-3:r^  +  4a:-2  3a5(a^-&0 

82^3-27^3       *  '       o?-^ah-ac      ' 

16.  o'-^'-^^o-o'  ^     y  26.     a'-(b  +  c  +  dy 


a'-{-2ab  +  b^-c'  (a  -  bf  -  (c  +  c^)^ 

a;^-a;^-23;  +  2  ^     6a^-5x-6 

'       2a;^-:r-l     *  *  8:^-2:i;-15* 


„    6rg^-23a;^  +  16ry-3  .^         :?;«  +  ?/ 


62;^- 17:^2+ 11:^-2  ^-^-^y^  +  y^ 

•  :r^_2a;«-:^-22^  +  l'  '  (a^-b')(a'-ab  +  b')' 

To  Reduce  a  Fraction  to  an  Integral  or  Mixed 
Expression. 

Change  — ^'— —  to  a  mixed  expression. 
X  —  1 

(a^+l)-^(x-l)  =  ar'-]-x  +  l  +  — ^.     Hence, 

x—l_ 

166.  7/"  ^Ae  degree  of  the  numerator  of  a  fraction  equals  or 
exceeds  that  of  the  denominator,  the  fraction  may  be  changed 
to  the  form  of  a  mixed  or  integral  expression  by  dividing  the 
numerator  by  the  denominator. 


FRACTIONS.  107 


The  quotient  will  be  the  integral  expression,  the  remainder 
(if  any)  will  he  the  numerator,  and  the  divisor  the  denom- 
inator, of  the  fractional  expression. 

Exeecis:e  LIII. 
Change  to  integral  or  mixed  expressions : 

,    x'-^x+l  ^    10aJ'-17ax-\-10a^ 

1. .  6. 

X—  1 

X-\-4: 

X  +  4L 
.    d^  —  ax-\-a^ 


7. 


8. 


a-\-x 


9. 


5. 


23^^  +  5  ^^ 

a;-3  *  ba^-\-4:X-l' 


ba  —  x 

16(3x^+1) 

4:X-1        ' 

2o^-5x-2 

x-4:    :   • 

a'  +  P 
a  —  b' 

ba^-x'-^b 

To  Reduce  a  Mixed  Expression  to   the  Form  of  a 
Fraction. 

167.    In  arithmetic  5|  means  5  +  1. 

But  in  algebra  the  fraction  connected  with  the  integral 
expression,  as  well  as  the  integral  expression,  may  be  posi- 
tive or  negative ;  so  that  a  mixed  expression  may  occur  in 
any  one  of  the  following  forms : 


,  a  a  ,  a  a 


108  ALGEBRA. 


Change  w  +  -  to  a  fractional  form. 

0 

Since  there  are  h  hths  in  1,  in  w  there  will  be  n  times  h 
hihs,  that  is,  nh  hths,  which,  with  the  additional  a  bths, 
make  nb-[-a  hths. 

a nh  -\-a 


In  like  manner 


^^+5      b 


a      nh  —  a 

n  — 


h  h      ' 

,  a      —nh-\-a 

-"+1= — h-'' 

1  a      —nh  —  a     TT 

and  —  n  —  -  = ; .    Hence, 

o  0 


168.  To  reduce  a  mixed  expression  to  a  fractional  form. 
Multiply  the  integral  expression  hy  the  denominator,  to 

the  product  annex   the   numerator,   and  under  the  result 
write  the  denominatx)r. 

169.  It  will  be  seen  that  the  sign  before  the  fraction  is 
transferred  to  the  numerator  when  the  mixed  expression  is 
reduced  to  the  fractional  form,  for  the  denominator  shows 
only  what  part  of  the  numerator  is  to  be  added  or 
subtracted. 

The  dividing  line  has  the  force  of  a  vinculum  or  paren- 
thesis affecting  the  numerator ;  therefore  if  a  minus  sign 
precede  the  dividing  line,  and  this  line  be  removed,  the 
sign  of  every  term  of  the  numerator  must  he  changed. 
Thus, 

a  —  h      en  —  (a  —  h)      en  —  a-\-h 

n = ^ ^= —. 

c  c  c 


FRACTIONS.  109 


(1)    Change  to  fractional  form  x—1- 

X 

_x^  —  x-{-(x—l) 

—  > 
X 

_a^~x-\-x—  1 

—  I 

X 

~       X 


,V1.\J1.X<XL 

X 

X  — 

-1  ^-\ 

X 

x"- 

-x-{x-X) 

X 

:^- 

-x-x-\-\ 

X 

7?- 

-2:r  +  l 

Exercise  LIV. 
Change  to  fractional  form  : 

1.  1      ^-y  K    ^^_QA  j3a2-452) 

x-\-y 

„    T    ,  x—y 

2.  1   '  ^ 


3.   3a;  — 


x-\-y 
1  +  2^:2 


X 

a'  +  x' 


5a-66' 

6. 

a  +  b     «^  +  f. 

7 

^7^      2-3a  +  4a2 

5-6a      ■ 

a 

Zx     ^'"-^. 

4.   a  — a: 

'    a  —  x  ''Za 


110  ALGEBRA. 


9.   ^+*  +  L  15.    2a-l-^. 

a  —  0  a-\-h 

10.    ^^-1.  16.    3a;-10  ■      ^^ 


a  +  b  ^  +  4 

11.  -^-(^  +  y).  17.   ^  +  a;  +  l  +  -?-. 

12.  5^Zll2^  +  6a  +  3^.  18.   .^_  3^,  3^(3 -^)^ 

4.  x~^ 

13.  a-l  +  — ^.  19.   o?-2ax-\-^x^        ^^ 


a+1  a-\-2x 

—  15  ^  ,       ,  a^  — «?/  + 


14.  ^  +  5-2^.  20.  *-a  +  y  '  ^'-'^V  +  v' 


Lowest  Common  Denominator. 

170.    To  reduce  fractions  to  equivalent  fractions  having 
the  lowest  common  denominator : 

3  T       2  77  5 

Eeduce  — -,  -^,  and  -— ,  to   equivalent  fractions  hav- 

ing  the  lowest  common  denominator. 

The  L.  C.  M.  of  4a^  3a,  and  Ga^  =  12al 

?>x 
If  both  terms  of  -— ^  be  multiplied  by  3  a,  the  value  of 

the   fraction  will   not   be    altered,  but  the"  form  will  be 

changed  to  — —  ;  if  both  terms  of  — ^  be  multiplied  by  4a^, 
lAcL  o  a 

O  ™2, . 

the  equivalent  fraction 4  is  obtained  ;  and,  if  both  terms 

12  a^ 

5  .    .  .  .        10    . 

of  -—  be  multiplied  by  2,  the  equivalent  fraction         ^  is 

obtained. 


FRACTIONS.  Ill 


ence,  — .    — ^.    • — 

1  i  9  ax     8aV       10  ,•     ^ 

areequalto  — ,   ^,    — ,  respectively. 

The  multipliers  3  a,  4  a^,  and  2,  are  obtained  by  dividing 
12  a*,  the  L.  C.  M.  of  the  denominators,  by  the  respective 
denominators  of  the  given  fractions. 

171.  Therefore,  to  reduce  fractions  to  equivalent  frac- 
tions having  the  lowest  common  denominator, 

Mnd  the  L.  C.  M.  of  the  denominators. 

Divide  the  h.  CM.,  hy  the  deTumiinator  of  each  fraction. 

Multiply  the  first  numerator  hy  the  first  qicotient,  the  sec- 
ond by  the  second  qicotient,  and  so  on. 

The  prodiLcts  will  he  the  numerators  of  the  equivalent 
fractions. 

The  L.  C.  M.  of  the  given  denominators  will  he  the  denom- 
inator of  each  of  the  equivalent  fractions. 

Exercise  LV. 

Keduce  to  equivalent  fractions  with  the  lowest  common 
denominator : 

,     3a:-7  4:X-9     y       g  1  1 


6      '      18  (a  -  b)(h  -  cy  (a  -  h){a  -  c)' 

2x  —  ^y  2>x—^y  g         ^a?  xy 

ba?     '      lOx    ■  •  3  (a +  5)'  6{a^-¥)' 

^a-bc  Sa-2c  ^  8a:  +  2  2x~l    Sx-{-2 

bac    '     12a2c  ■  '  x-2' 2>x-6' bx-lO' 


.        5  6  a  —  hm  -,    c  —  hn 

I—  X   i  —  XT  mx  nx 


112  ALGEBBA. 


Addition  and  Subtraction  of  Fractions. 

172.  To  add  fractions : 

Reduce  the  fractions  to  equivalent  fractions  having  the 
lowest  common  denominator. 

Add  the  num^erators  of  the  equivalent  fractions. 
Write  the  result  over  the  lowest  common  denominator. 

173.  To  subtract  one  fraction  from  another : 

Reduce  the  fractions  to  equivalent  fractions  having  the 
lowest  common  denominator. 

Subtract  the  numerator  of  the  subtrahend  from  the  numer- 
ator of  the  m^inuend. 

Write  the  result  over  the  lowest  common  denominator. 

(1)  Simplify,       4£±7^3x-4^ 

5  15 

The  lowest  common  denominator  (L.  C.  D.)  =  15. 

The  multiphers  are  3  and  1  respectively. 
12  a;  -f-  21  =  1st  numerator, 
3  a:  —    4  =  2d  numerator, 
15a;  +  17  =  sum  of  numerators. 
.    4a;  +  7   ■    3a;-4^  15a:+17 
"       5       "^     15  15      • 

/ox    o-      T-c         3(2  — 4 &      2a  —  b-{-c,lSa  —  4:C 

(2)  Simplify, 3~  + 12 * 

The  L.  C.  D.  =  84. 

The  multipliers  are  12,  28,  and  7  respectively. 

36  a  —  48  &  =  1st  numerator, 

— 56  a  +  28  5  —  28  C  =  2d  numerator, 

91a —  28  g  ==  3d  numerator. 

71a  —  20  b  —  56  0  =  sum  of  numerators. 

•    3a-45       2a~b  +  c       13a- 4c  _  71a- 20^>  -  56g 


12  84 


FRACTIONS.  113 


Since  the  minus  sign  precedes  the  second  fraction,  the  signs  of  all 
the  terms  of  the  numerator  of  this  fraction  are  changed  after  being 
multiplied  by  28. 


Exercise  LVI. 
Simplify : 

bx      "^      10a;      "^       25      ' 

^x^-7i/      3x-8y      5-2y 
•        3ar^       "^       6:r       "^       12     • 

2^2       "^       55       "^      9     * 


.    4a;  +  5       3a:-7    ,      9 
4.    — z r 


bx  Ux" 

g    4a;-3y       3a:  +  7y_(5a;-2^    (9:^  +  2^) 
7        "^       14  21       "^       42     * 


6    ^^y~^  _  53/'  +  7  _  e^r^-ll 
afi/^  xy^  a^y 


o    5^3-2       3a2-a 

o«     r — :: z . 


o    a  —  b.h  —  c,c  —  a,    alP'  -\-h^-\  cg? 
9. 1 1 1 . 

cab  abc 


'   2ary       6yh       2xr        4:iV  ^   4:x'yz' 


114  ALGEBRA. 


Simplify  ^      y  +  ^±^. 
x-\-y      x  —  y 

TheL.C.D.  =^-3/l 

The  multipliers  are  x  —  y  and  a;  +  y  respectively. 

Of?  —  2iXy  -\-    y^  '■=■  1st  numerator, 
x^ -{-^xy -{-    2/^  =  2d  numerator. 
2;z^  +  23/^  =  sum  of  numerators, 

or,  2{x'  +  f)  =    "    " 

"  x  +  y      x-y         x^  —  f 

Exercise  LVII. 
Simplify : 


o;  —  6      a;  +  5'  2 a  (a  +  ^)       2a  {a  — x) 


11  „  a  h 


X 


—  7       x  —  Z'      ■  {a  +  b)b       {a  —  h)d 

1.1  «  5  3 


l  +  o;       1-a;  2:i;(a:-l)       4:x{x-2) 

^    _1 2_  g         l  +  a: 1-07 

'    1—x       l  —  a?'  '    l-\-x  +  a^       1—x  +  x^' 

1  a;  ^     2(2:r  — 3^3/        2aa;H-35?/ 

^-2/      W—yf  '   2xy{x  —  y)       2xy{x  +  y)' 

r-iN    o-      vr     2a-\-h       2a  — h         ^ab 
(1)    Simphfy-^--^--,-^. 

TlieL.C.D.  =  (a-5)(a  +  5). 

The  multipliers  are  a  +  6,  a  —  6,  and  1 ,  respectively. 


FRACTIONS.  115 


2a^  -{-Sab  -{-  P  =  1st  numerator, 

2a^ -{-  S ab  —  b^  =  2d  numerator. 

—  6ab  =  3d  numerator. 


0  =  sum  of  numerators. 

.   2a-\-b       2a  — b         6ab    _^ 
' '    a  —  b         a-{-b        a^  —  b^ 

(2)   Simplify  -J^,-^+l  +  -|M^.       . 

The  L.  C.  D.  -  (2:  +  y)  (:r  -  y)  (:r2  +  y^). 

The    multipHers    are    a^  +  y*,     {x  —  y){^  -^  y'),     {x  +  y)  (x  —  y) 
ipi?  +  y'),    {x  +y){x  —  y),  respectively. 

^^  +  y*  =  1st  numerator, 

—  x^-}-  23?y  —  2o^\^  -\-  2x\f  —  y*  =  2d  numerator, 

x^  —  y*  =  3d  numerator, 

2  :i?y  —  2  xif  =  4th  numerator. 

4  ^y  —     x^y^  —  y*  "=  s^°^  of  numerators. 

4  3^1/ x^i/  ?/^ 

.*.  Sum  of  fractions  =  — ^— r — ^^ — ^. 

x*-y* 

Exercise  LVIII. 
Simplify : 

1       ,       1       ,      2<z            o       ^             x^      ,       X 
1-    :r-^ r  z h  :; o-        3. — h 


1  +  a       1  —  a       1  —  a^'  1  —  :r       1—x       l-\-3r^' 

_i L_  +  _^.         4.   f  +  -^  +       ^ 

1  —  iC       1  +  :r       1  -f^a;^  y       a;  +  y       a:^  +  a;y 

x-^2'^  x-S'^  x  —  4: 
^       3       ,       4a  5a2 

b. 1- 


x  —  a       (x  —  ay       (x  —  ay 


116  ALGEBEA. 


7.        1  ^ 


X 


-1       rr  +  2       (a;+l)(^  +  2) 


a  —  h  ,  h  —  c ,  G  —  a 

"•   ~r, — ; — z~} — : — ^   i    ; — ; — tt — ; — t\  ~r 


(5  +  c)(^  +  a)  {c  +  a){a-\-b)       {a -\- h)  {h -{- c) 

x  —  ax—h  (a  — by 

x—h       X  —  a  {x  —  a)(x  —  b) 

'    {b-c){c~ay  {c-a){a-by    {a-b){b-c) 

g}  —  be ,  W  —  ac  ,  c^  -{- ab 

'    ^a  +  b){a  +  cy  (b-\-a){b  +  c)'^  {c-\-b){c  +  a)' 

a  X      .  c?-\-0(? 


a  —  x       a-{-2x       (a  —  x)  (a-\-2x) 
14.    ,         ±         ,  -  ^^ ,+ 


(a  —  b)(b  —  (?)       (a  —  b){a  —  c)       (a  —  c)(b  —  c) 

x  —  2y  2x-\-y  2x 

x{x  —  y)       y{x  +  y)       3?~y^' 


^^       a  —  b a  —  b      _  {a  —  b)  {x  +  y) 

'   x{a  +  b)       y{a  +  b)  xy(a-\-b) 

17        Brr  :^  +  2y  3y 


18. 


(x  +  yf       x'-f       {x-yf 

a  —  c a  —  b 

ia  +  bf-c"       ia  +  cf-b^' 


19.     ^  +  ^    _    ^~^     t   ^^ (^  —  y) 

ax  -\-  by       ax  —  by       c^oi?  —  b^if 


FRACTIONS.  117 


174.    Since  %-  —  a,  and =  a, 

h  —0 

it  is  evident  that  if  the  signs  of  both  numerator  and  de- 
nominator be  changed,  the  value  of  the  fraction  is  not 
altered. 

a  —  b       —{a  —  h)       —a-\-b       h—a 


Again, 


c  —  d      ~{c  —  d)       —c-\-d 


Therefore,  if  the  numerator  or  denominator  be  a  com- 
pound expression,  or  if  both  be  compound  expressions,  the 
sign  of  every  term  in  the  denominator  may  be  changed, 
provided  the  sign  of  every  term  in  the  numerator  be  also 
changed. 

Since  the  change  of  the  sign  before  the  fraction  is  equiva- 
lent to  the  change  of  the  sign  before  every  term  of  the 
numerator  of  the  fraction,  the  sign  before  every  term  of 
the  denomiyiator  nnmj  be  cliangedy  -provided  the  sign  before 
the  fraction  be  changed. 

Since,  also,  the  product  of  +  «  multiplied  by  +  ^  is  ab, 
and  the  product  of  —  a  multiplied  hj  —b  is  ab,  the  signs 
of  two  factors,  or  of  any  even  number  of  factors,  of  the  de- 
nominator of  a  fraction  may  be  changed  without  altering 
the  value  of  the  fraction. 

By  the  application  of  these  principles,  fractions  may  often 
be  changed  to  a  more  simple  form  for  addition  or  subtrac- 
tion. 

(1)    Simplify2_       3  2^ 

Change  the  signs  before  the  terms  of  the  denominator  of  the  third 
fraction,  and  change  the  sign  before  the  fraction. 
The  result  is, 

2_       3  2.T-3 

X      2x-l      4ar^-l' 

in  which  the  several  denominators  are  written  in  symmetrical  form. 


118  ALGEBRA. 


The  L.G.D.=x(2x-l)(2x+  1). 

Scc^  —  2     =:  1st  numerator, 

—  6a^  —  Sx=^2d  numerator, 

—  2x^-j-  3x  =  3d  numerator. 

—  2     =  sum  of  numerators, 

-2 


'.  Sum  of  the  fractions  = 


a;(2a:-l)(2a7  +  l) 
(2)    Simplify 

1 


+  V-7-. ^T^ ^  + 


a(a~h)(a  —  c)       h  (b  —  a)  (h  —  c)       c  (c  —  a)  (c  —  h) 

Change  the  sign  of  the  factor  (6  —  a)  in  the  denominator  of  the 
second  fraction,  and  change  the  sign  before  the  fraction. 

Then  change  the  signs  of  the  factors  (c  —  a)  and  (c  —  h)  in  the  de- 
nominator of  the  third  fraction. 

The  result  is, 

111 


a{a  —  h){a—c)       b(a  —  h)  {h  —  c)       c(a  —  c)  (b  —  c) 

in  which  the  factors  of  the  several  denominators  are  written  in  sym- 
metrical form. 

The  L.  C.  D.  =  ahc  (a-h)(a-c)(h  -  c). 

he  (b  —  c)  =  Wc  —  bc^  =  1st  numerator, 

—  ac  (a  —  c)  =  —  o^c  -f-  ^c^  =  2d  numerator, 
ah  {a  —  ^)  =  c^b  —  aV^               =  3d  numerator. 

o^b  —  0?G  —  ab"^  +  a(?  +  ^^'^  —  bc^'^^  sum  of  numerators. 
=  d?{h-c)-a  (b''  -c')-\-bc(b-  c), 
=  [a^-  a(b  +  c)  +  bc][b  -  c], 
=  [a^  —  ab  —  ac~\-  he]  [b  —  c\ 

-  [(a^  -  ac)  -  {ah  -  he)]  [h  -  c], 
=  [a{a  —  e)  —  h(a  —  e)]  [b  —  c], 
=  (a  —  b)(a  —  e)  (h  —  c). 


FRACTIONS.  119 


Sum  of  the  fractions  =  -7^^^ ^-^rj ^\tl — ^ 

aoc  {a  —  o){a  —  c)  [p  —  c) 

^_1^ 

abc 


Exercise  LIX. 
Simplify : 

X  X  —  y 


x  —  y      y  —  X 

3  +  2a:   ,    337-2   ,    \^x-x^ 
2~x         2  +  x  "^    .^2-4   * 

X^  I  X  X     .- A^^ 


rr*  —  1       x-{- 1       1  —  a: 
4.       4^,+  „_JL_+      1 


V 


3-82/*       2-2?/       6y  +  6' 

1  2.1 


(2-m)(3-m)       (m-l)(m-3)       (m-l)(m-2) 
1.1  ' 


+ 


(Z>-a)(a;  +  a)       {a-h){x  +  b) 


rv  ->  A 


Q    5  —  a  a  —  25  3.'r(«  —  5) 

x  —  b  b-\-x  W  —  x^' 

g    3  +  2a?       2-3a;  .    IS'rc-a:' 

2-a;         2  +  a;  a:2-4  * 


3 7  4-20a: 

'    1-2^       l  +  2:c        4a;2_|- 


120  ALGEBRA. 


, ,  a  +  b I b-\-c I c-\-a 

{b  —  c){c  —  a)       {b  ~a){a  —  c)      {a  —  b){b  —  c)' 

12  c^  —  bc         ,  b^  -\-ac  ,  c^  -{- ab 

{a-b){a-c)       (5  +  c)(6-a)"*'(c-a)(^  +  ^)' 

13.         y+g        J         ^+^        J       '^+y 


14. 


(x-y){x-z)       {y-x){y-z)       {z-x){z-y) 

3 4 6 

{a  —  b){b  —  c)       (b  ~a){c  —  a)       {a  —  c){c  —  b) 


15.    , \ +  1 


x{x-y){x-z)       y{y-x){y-z)       xyz 


Multiplication  of  Fractions. 

175.  Hitherto  in  fractions,  equal  parts  of  one  or  more 
units  have  been  taken.  But  it  is  often  necessary  to  take 
equal  parts  oi  fractions  of  units. 

Suppose  it  is  required  to  take  -|  of  -|  of  a  unit. 

Let  the  line  AB  represent  the  unit  of  length. 

^  I    I    I    I    I    I    I    I    1    I    I    1    I    I    I    l-g 
C  D  E  F 

Suppose  AB  divided  into  5  equal  parts,  at  C,  D,  E, 
and  F,  and  each  of  these  parts  to  be  subdivided  into  3 
equal  subdivisions. 

Then  one  of  the  parts,  as  A  (7,  will  contain  3  of  these  sub- 
divisions, and  the  whole  line  AB  will  contain  15  of  these 
subdivisions. 

That  is,  -I  of  -J-  of  the  line  will  be  -^  of  the  line  ; 

-1  of  I  will  be  J^  +  ^V  +  tV  +  t^'  or  •^,  of  the  line  ;    and 

I  of  I  will  be  twice  3^,  or  -j^,  of  the  line. 


FRACTIONS.  121 


Suppose  it  is  required  to  take  -  of  -  of  the  line  AB. 


><l    t    I    I    I    I    I    I    I    I    I    I    I    I    I    17? 
G  D  E  F 

Let  tlie  line  AB  be  divided  into  h  equal  parts,  and  let 

each  of  these  parts  be  subdivided  into  d  equal  subdivisions. 

Then  the  whole  line  will  contain  hd  of  these  subdivisions, 

and  one  of  these  subdivisions  will  be  — -  of  the  line. 

hd 

If  one  of  the  subdivisions  be  taken  from  each  of  a  parts, 

they  will  together  be  ^  of  the  line.    That  is, 
hd 

¥l  =  h^h'^h takenaWs,  =  ^^, 

and  -  of  -  will  be  c  times  --_,  or  -—  of  the  line. 
d      b  hd       hd 

Therefore,  to  find  a  fraction  of  a  fraction. 

Find  the  product  of  the  nuTnerators  for  the  numerator  of 
the  product,  and  of  the  denomiruxtors  for  the  denominator 
of  the  product. 

176.    Now,  -  X  -  means  -  of  -. 
d      h  d      h 

Therefore,  to  find  the  product  of  two  fractions, 

Find  the  product  of  the  numerators  for  the  numerator  of 

the  product,  and  of  the  denominators  for  the  denominator 

of  the  product. 

The  same  rule  will  hold  when  more  than  two  fractions 
are  taken. 

If  a  factor  exist  in  both  a  numerator  and  a  denominator, 
it  may  be  cancelled ;  for  the  cancelling  of  a  common  factor 
before  the  multiplication  is  evidently  equivalent  to  cancel- 
ling it  aftey:  the  multiplication ;  and  this  may  be  done  by 
§  163. 


122  ALGEBRA. 


Division  of  Fractions. 

177.  Multiplying  by  the  reciprocal  of  a  number  is  equiv- 
alent to  dividing  by  the  number.  Thus,  multiplying  by  i 
is  equivalent  to  dividing  by  4. 

The  reciprocal  of  a  fraction  is  the  fraction  with  its  terms 
interchanged. 

Thus,  the  reciprocal  of  f  is  f ,  for  f  X  f  =  1.  §  42. 

Therefore,  to  divide  by  a  fraction, 

Interchange  the  terms  of  the  fraction  and  multiply  hy  iJie 
resulting  fraction.     Thus : 

2a    .     1   ^  2a       2>x  ^2a 

The  common  factor  cancelled  is  3  a;. 

14^  _^1x^  14^  X  ?^  =  ?^ 
^  ^    273/2  •   9^       21  f      1x       Zy 

The  common  factors  cancelled  are  9  3/  and  Ix. 

„  ax      _^      ah ax^ (a  +  :r)  (a  —  x) 

(a  ~  xy   '   0?  —  x^       {a  —  x)  {a  —  x)  ab 

_x{a-\-x) 
b  (a  —  x) 
The  common  factors  cancelled  are  a  and  a  —  x. 

If  the  divisor  be  an  integral  expression,  it  may  be  changed 
to  the  fractional  form.  §  162. 


Exercise  LX. 

1. 

f.x 

ex 
d' 

3. 

Sp 
2p-2 

•  p-1 

2. 

2x 
—  X 
a 

Sab 
c 

Sao 
^  2b' 

4. 

8x*i/ 
Ibab^  ' 

2x^ 
Sab'' 

FRACTIONS.  123 


g     8a^b^        iDxy"  g     9mV       bfq       l^x'y' 

*  10 a'b'c  ^       IS x2/z'  '    UnY       TS^m       Ak'n 

^    3^  ^  V!  ^  _  12^  ^^      a-b   ^  a?-h\ 

4iX2^       6x1/           2x7^'  '   o?-{-ab      a?  — ah' 

a^  +  b''  _^  a-h  cc'  +  x~2       a:^-13ar  +  42 

*  oj'-b^   '  a  +  6'  ^-"^x             x'  +  2x.      ' 


x^~6x-\-2        x^  —  bx  o?-\-s(?      {a  —  x) 

15.    ^ct{:^-,ff  ^  ^ 

ex  {x  —  y){x-\-yf 

^^    a^^2ah  ^  ah -2b''  ^^   a?  +  xy  ^  {x-yf 

a^  +  W        G?—^h^'  x  —  y         x^  —  y*' 

17      ^  —  ^    Y  ^~~^^  iq    ^^  —  '^^^  _^  n  —  m 

'   a^  +  bx      x'  +  2x  '    (^  +  d^    '    c  +  d' 

a^-4a  +  3        a^-da  +  20       a^-7a       . 
'   a2-5«  +  4      a^-10a  +  2l       a^-ba 

b'-7b  +  6      P+lOb  +  24:  _^  F^  +  6h 
'   b'  +  Sb-4:      b'-Uh  +  48  ■   b^-Sb^' 

22.         ^-y^        X  ^'^~^^^  X  ^~^    ' 

:r2-3:ry  +  22/2        a;2_|.^y        {x-yf 

a^~-?,a?b-\-?>aW-b^  .   2a5-2Z>^      a^  +  a5 
a=^-62  *  3  a-6' 


124  ALGEBRA. 


24. 


d'-ib-cf    '   c'-la-bf 


{x-af-lP      c^-il-df 
•    (^:c-bf-o?      x'-ia-bf 


26. 


{a  +  cf~{b-{-df  '   {a-bf-id-cf 


^^   x^-2xy-^f-z^  ^  x  +  y-z^ 
cc^-^2xy-{-y^  —  z^       x  —  y-^z 

Complex  Fractions. 

178.  A  complex  fraction  is  one  which  has  a  fraction  in 
the  numerator  or  in  the  denominator,  or  in  both. 

179.  A  fraction  may  be  regarded  as  the  quotient  of  the 
numerator  divided  by  the  denominator. 

This  is  the  simplest  meaning  of  a  complex  fraction. 
Therefore,  to  simplify  a  complex  fraction, 
Divide  the  numerator  by  the  denominator. 

(1)    Simplify  i. 


1 

i-^i  = 

ixi- 

=  i- 

(2) 

Simplify  -5. 

^8 

■^ 

2|_  1  . 

5|     ¥' 

=  f-^¥ 

=  |x 

A  = 

T^- 

(3) 

Simplify  -^. 

Zx             ^x 

X-\          4:X~1 

_3x  . 
1 

4 

__%x 

1 

-it 

■1 

4 

12a; 

- 

4a;-l 


FRACTIONS.  125 


It  is  often  shorter  to  multiply  both  terms  of  the  fraction 
by  the  L.  C.  D.  of  the  fractions  contained  in  the  numerator 
and  denominator. 

Thus,  in  (1),  multiply  both  terms  by  6  ;  in  (2),  both  terms 
by  24 ;  in  (3),  by  4.      The   results   obtained   are  |,   j^^, 

\2x 

respectively. 


4:X-1 

(4)   Simplify 


1  +  ^ 


l-x  +  x" 


x(l  —  x-\-x^) 


1    1^1  ^  {l+x)(l-x-{-x')-i-x 


-I  *L  ju    ~]~  ju 


l+X  +  T? 

x{l-\-x^a^) 


l-\-x-\-a^-{x~a^-\-a?) 
_  X -\- 01? -{- x^ 

The    expression    is    reduced   to   the 

1  +  ^  +  T-^— 5 
form  x(l-x-\-:^) ^^^^^  _  x-x^  +  a?^ 

{l-\-x){l-X  +  C(?)+X  l-\-X-^3? 

The  expression  ■ — is  reduced  to  the  form 

1       x  —  or-{-cr 

l-{-x-\-a? 
a;(l  +  a;  +  :^)  which  -  ^  +  ^  +  ^'' 


l+x  +  x^-i^x-x'  +  a?)  l  +  x" 


126  ALGEBRA. 


EXEBCISE   LXI. 

Simplify : 

Sx  ,   X  —  1 

8.    1--1 


2 

'       3 

13 
6 

(.-M)-f- 

-2i 

X  — 

-1-f 

6 

^-6 

X  — 

-2-1 

3 
^-6 

3 

2x- 

1 

l+i 

X 


2.  :___:  9.  1 


1  +  .;+  ^"^ 


^+1      ^+5_i  1- 


1-x 

1 
10. 


1 


2     2  1  +  1 

X 

4.    ^^L£ 11.   L 

.  ^  _  (^  -  ^)  (^  -c)  -,  _, ^ 


^  +  «  1  +  0;+    2^ 


1- 


a; 


f-_-)f-+-)         f-+-_2y^+f+2) 

^    \x      aj\x      a  J  ^^    \x      a        J\x      a        J 

/a  __  ^Y 
\x      a) 


1  __x  —  a 

x-\-a 

1 ^ 

xj—yo^^  —  jf  x—4:-\ ^         1 


^  +  5 
y  13.   -^-  X 


xy-\-f     a^  +  xy  x 

x  —  1 

•      ^±1,^-1  {x'-f){2x'-2xy) 

^    ;r-l"^a;  +  l  ^^             ^{^-yY 

37+l_37  — 1  '                  ^y 

;r  —  l:i;+l  ^'-fy 


(a;-l)(a;-2) 


FRACTIONS.  127 

ab ac 

3(^ -\- {a -\- h)  X -\- ah      a^ -\- {a -\- c)  x -\- ac 

15.    _ 

b  —  c 


u  —  V 

x^-\-{b-{-c)x-\-bc 

a-\-b  . b_ 


16.    ^-  +  1 i_-      ^  17.    J ±±* 

1  +  i  -  +  1  i  +  i 

:r  a      0 

2m-3  +  -  J_  +  i  +  JL 

18.    VI       19.  ^^      ^"      ^"       20.   ? 

277^-l  a^-^b  +  cf  ^      3_ 

m  ab  -14^ 


l-x 
Exercise  LXII. 

miscellaneous  examples. 

2.  Find  the  value  of  ^'  +  f  ~^  +  ^^^  when  a  =  ^,b  =  i, 

_  1  a^  —  6^  —  dT  +  2  ic 

2ai^      ^3 

3.  Find  the  value  of  Sa^  H ^ 7^  when  a  =  4,  Z>  =  J, 

c  =  l.  ^-        ^ 

4.  Simplify  — -? — — i ?— . 

X  X 

6.  Find  the  value  of  f ^^Y-  £zi2a  +  5  ^^^^  ^  =  ^L±A^ 

\x  —  bj      x-{-a  —  2b  2 

7.  Simplify  \^^~  -  -^^^:^-  +  -?^  1  ^i::!^. 

^    M2(a~^)      2(a  +  bya'-b')    2b 


128 


ALGEBRA. 


8.  Simplify  [plt^  - 1^  ^  to  _  ^\ 

\:x?-y^     x'^-yV      \x-y      x  +  yj 

9.  Simplify 

\f~   J  \^-y        J      \f        J  [x'  +  xy  +  y^ 
10.    Simplify 

fa'-ab'Kfa^  +  ab  +  h^  ,  (^2a^  _  i  Vl  -  - 
\o?-by\      a  +  b       y\a?-^rb^        J\        a?-\-ab-\-b\ 


2ab 


11.   Simplify 


1  + 


^      a-~x    '    ^       d^  —  o? 


a  —  X 
a-\-x 


a-\-x  a^-\-o^ 

12.   Divide  a^J^^-2>C^-J\  +  4:(x-{-^hYx-\-^. 
1  2:ry 


13.    Simplify 


{x  +  yY 


2xy 


{x  -  yj 


1+' 


X} 


x-\-  2a  ,  X—  2a 


4:ab 


when  X 


14.  Find  the  value  of     ,  ,      ,  „       „ 

^^  2b~x      2b-\-x     U^-x" 

^a  +  b' 

15.  Find  the  value  of  ^  +  y~}  when   x  =  -^^t-^    and 

_ab  +  a  ^-y+1  «^  +  l 

^    ab+r 

16.  Simplify 
1 


+ 


+ 


1 


7- 


a{a  —  b){a~c)       b(b  —  c)  (b  —  a)       c{p  —  a)  {c  —  b) 
Sabc 


a-\      b-\      c-1 


17.   Simplify 


bG-\-ca  —  ab 


a      0      c 


FRACTIONS.  129 


^  2  2 

18.    Simplify X     3  ,     3 

n     m 


19.  Simplify   ^— *±£|i  +  ^!+^z:±'| 

20.  Simplify    3a-[6  +  12a-(^-c)n  +  |  +  |^ 

1        1 L    ^  y 


21.  Simplify  g-^-     a-.v     («-^y     (a-.y)^ 

(a-y)(a-a;)2      (a-a:)(a-y)2 

22.  Simplify    ^ 23.    (^-.V^)  (2^^- 2.,/) 

3-a; 

24.  Simplify    ^^Zl^  -  ^:\  ^  f  ^  +  ^^ 

25.  Simplify  ?/  |  ^  j  -^+y 

(:c-2/)0r-2)      (y-a;)(y-2)      {z-x){z-y) 

a;-4  +  -^         1-^+^ 

^r4-l  9'2_-i 

27.    Simplify    ^-^-1^  X 


""      x-\ 


CHAPTER  IX. 

Fractional  Equations. 
to  reduce  equations  containing-  fractions. 

180.   (1)  ^  4-  ^  =  12. 

2      4 

Multiply  both  sides  by  4,  the  L.  C.  M.  of  the  denominators. 
Then,  2a;  +  a;  =  48, 

3a;  =  48, 
.-.  a;  =  16. 


(2)  2-4  =  24- 

X 

8* 

Multiply  both  sides 

by 

24,  the  L.  C.  M.  of  the  denominators. 

Then, 

4a; -96  =  576 -3a;, 
4a;  +  3a;  =  576 +  96, 
7a;  =  672, 
.-.a;  =  96. 

(3)^           ^-1::= 
^^     3                11 

X  - 

-9. 

Multiply  by  33,  the  L.  C.  M.  of  the  denominators. 

Then,  lla;-3a;  +  3  =  33a;~297, 

11  a;  -  3a;  -  33a;  =  -  297  -  3, 
-25a;  =  -300, 
.-.  a;  =12. 

Since  the  minus  sign  precedes  the  second  fraction,  in  r,emoving 
the  denominator,  the  +  (understood)  before  x,  the  first  term  of  the 
numerator,  is  changed  to  — ,  and  the  —  before  1,  the  second  term  of 
the  numerator,  is  changed  to  +. 

181.    Therefore,  to  clear  an  equation  of  fractions, 
Multiply  each  term  hy  the  L,  C.  M.  of  the  denominators. 


FRACTIONAL  EQUATIONS,  131 

If  a  fraction  is  preceded  by  a  minus  sign,  the  sign  of 
every  term  of  the  numerator  must  he  changed  when  the 
denominator  is  removed. 

Exercise  LXIII. 
Solve  the  equations : 

1.  5x-^-±^  =  n.  4.  ^-^=?-^:z^ 

2  2        4       4        2'  * 


,    5-2a:  ,  o_         Gx-8    _    :r  +  2_14      S  +  bx 
o»    : r  ^  —  X — .  o.    — - — •  —  — . 

4  2  2         9  4 

^    5x  +  S      S-4:x  ,  X      31      9-5x 


+ 

=  10(a;-l). 


8  3*22  6 

10^+3  __  6^— 7 
3  2 


9.   5^:z7_2^±^^3^_i4. 
2  3 

10    7a;  +  5      bx-6_^8-6x 
*        6  4  12    * 

13.  i  (3a;  -  4)+ i  (5a; +  3)  =  43 -5a;. 

14.  i(27-2a;)  =  |-i(7a;-54). 


132  ALGEBRA. 


15.   5x-{8x-S[16-Qx-(i4:-5x)]l  =  e>, 

5x-S      9-x_6x  ,  19 

7  3  2  "^  6 

2^+7      9:r-8._a;-ll 


16.    5a:-3_9-:^^5^      19.    _^. 
7  3  2        6^         ^ 


"•        7  11             2     • 

_o    82^-15  lla;-l_7a;  +  2 

3  7              13 

7;r  +  9  3a;+l_9^-13      249-9a; 

'       8  7                4                 14      * 


182.  If  the  denominators  contain  both  simple  and  com- 
pound expressions,  it  is  best  to  remove  the  simple  expressions 
first,  and  then  each  compound  expression  in  turn.  After 
each  multiplication  the  result  should  be  reduced  to  the 
simplest  form. 

,,^  8x  +  5  ,  7x-S  _4:x  +  6 


'^'      14      '  6x  +  2 

7     • 

Multiply  both  sides  by  14. 

Then,                 8a;  +  5  +  1^1:=^  =  8a;  +  12. 
3a;  +  1 

49  a;      '^l 
Transpose  and  combine,  — ~~  =  7. 

Multiply  by  3  a; +  1, 

49a; -21  =21a;  +  7, 

28a;  =  28, 

.-.  a;  =  1. 

3-4^            !^-3 
^^4          4    ■     10 

Simplify  the  complex  fractions 

1  by  multiplying  both  terms  of  each 

,ction  by  9. 

Then,              27-4a;^ 
36 

1      7a;  -  27 
4          90     ' 

Multiply  both  sides  by  180. 

135 -20  a;: 

=  45 -14a; +  54, 

-6a;: 

=  -36, 

.*.    X- 

=  6. 

FRACTIONAL  EQUATIONS.  133 


Exercise  LXIV. 
Solve  the  equations : 

36  5^-4"'^4' 

9(2rr-3)  ■  lla:-1^9:r  +  ll 
14       ""^32:+l  7      • 

10^  +  17       12^7  +  2  __5a:-4 
18  13a:- 16  9      ■      J 

6a:  +  13       2>x  +  b  _2x 
15  5a: -25       5* 

„     18a:-22,o     ,l  +  16a:_.5       101-64a: 
'•    39^^67  +  ^''  +  ~2^-^^  24^- 

6    C-5a7        7-2ar^    _l  +  3a:      10a:-ll        1 
'       15  14  (a: -1)  21  30  105* 

9a:  +  5  .  8a:- 7  _36a:+ 15  ,  41 

*  14     '^6a:  +  2  56       "^56*" 

6a:  +  7      2a:-2__2a:  +  l 

*  15     "    7a:-6  5      * 

6a:+l       2a:-4  _2a:-l 

*  15  7a: -16"        5     ' 


10. 


7a:  —  6         a:  —  5 


__x 

35    .     6:1- -  101  ~  5' 


183.  Literal  equations  are  equations  in  which  all  the 
numbers  are  represented  by  letters ;  the  numbers  regarded 
as  known  numbers  are  usually  represented  by  the  first  let- 
ters of  the  alphabet. 


134  ALGEBRA. 


(1)  (a-x){a-\-x)  =  2a^-\-2ax  —  3i?. 

Then,    a? -x"  =  20^ -V  2ax-x^, 
—  2ax  =  a^, 

a 
.\  x  = . 

2 

(2)  (x  —  a)  (x  —  h)  —  {x  —  h)  (x  —  c)  =  2(x  —  a)  (a~c). 

{x^  —  ax  —  bx  +  ah)  —  {x^  —  bx  —  cx  +  bc)  =  2  {ax  —  cx  —  a^+  ac), 
a^  —  ax  —  hx  +  ab  —  x^  +  bx  +  cx  —  bc==2ax  —  2cx  —  2a^  +  2ac. 
That  is,  —3ax  +  3cx  =  —  2a^  +  2ac--ab  +  bc, 
—  3  (a  —  c)  j;  =  —  2  a  (a  —  c)  —  6  (a  —  c), 
-3x  =  ~2a~b, 
_2a  +  b 

Exercise  LXV. 
Solve  the  equations : 
1.    ax  -\-  be  =  bx  -{'  ac.  2.    2a  —  cx^=^c  —  b  bx. 

3.  a^x  -\-bx  —  c  ^=^Wx  -\-  ex  —  d. 

4.  —  ac^  +  b'^e  -j-  obex  =  abe-^-  emx  —  ae^x  +  b^e  —  me. 

5.  (a  +  :r  +  S)  (a  +  S  —  a;)  =  (a  +  a;)  (b  —  x)  —  ab. 

6.  (a2  +  rr)2  =  :r2  +  4a2  +  a^ 

7.  {a^-x){a^-{-x)=-a'^-\-2ax-x^. 

.ax  —  b,  x-}-ac  ^-  Sa  —  bxl 

9.    ^il^+^  =  a.:.+  ^.  11.    Ga-^^'^-^^^x. 

bx  b  3 

a^  —  a      a  —  x      2x      a 


12. 


5a;  b  b       X 


3      a5  —  a;^)_4:r  —  ac 
c  ^07  ex 

14.    am  —  &  —  -=--] =  0. 

o       m 


FRACTIONAL  EQUATIONS.  135 

^ax  —  2  b      ax  —  a «£_2 

•   ~    3b  2b     ~  b       3' 

ifi    ab_-\-_x_P  ~  X x~b      ab  —  X 

'       b'  a'b    ~    oT  b'     ' 

bx-{-l      a(x^—l)  ^r.    oh      7.7,1 

Vt.   ax —  =  -^^ -.  19.    ~=bc  +  d-\--. 

la   _^  +  a  +  ^  =  0.  20.   "(^'  +  ^)^ac  +  f. 

0  —  cx  c  ax  a 


Exercise  LXVI. 
Solve  the  equations : 

-        a:  — 3     __    x  —  f)      .  1 

4:{x-l)~Q{x-iy^ 

2.   x  I      ^     ^_(:^— 2)(a;+4)^ 
x  —  1  x-{- 1 

3        7      ^6:r  +  l      3(l  +  2:g^) 

'     X~l  X-\-\  3?  —  1 

__J^ 1,        _  x-\ 

2{x-3)      3(a:-2)      {x-2){x-Z^  1 

g    .      2(2a:  +  3)_     6  hx-^\ 

9(7-x)       7-x     AO-x)' 

6.   _iL_4  =  ^i?l  +  2^_10. 
a;  +  3  3:r  +  9 

^    x-7      2:r-15  1 


x  +  7       2a; -6       2(x-\-1) 
3a:  +  5"^   ^      2a;  +  3' 


136 


ALGEBRA. 


9. 


10. 


132;r-M   ,  8x  +  5 


Sx-i-1 
2 


x-1 


52. 


11. 


12. 


3x-l      Ax- 

-2      1 

2a;-l      ^x- 

-2      6' 

3          :^^+l_ 

x' 

2x-S      x-2      Sxi-2  x-1 

x  —  4:     x  —  5      x  —  7     x  —  8 


1      1-^' 


y 


13. 


x—b      x—6      x—8      x—9 

\2 


14.  {x  —  a)(x  —  b)  =  (x  —  a  —  by. 

15.  (a—h)(x—c)  —  (b  —  c){x~a)—(c—a)(x—h)=0. 


16. 


17. 


a::^  —  :^+l    ,   x^-}-  x-{-l 


1 


2:r. 


37 


^  +  2     x  +  S     s^  +  dx  +  Q>' 
18.    (a;+l)2  =  :r[6-(l-r^;)]-2. 


19. 


20. 


21. 


25-^x      Ux  +  Aj 


23 


x+l 
Sabc   , 


3ri;  +  2         :r  +  l 


4-5. 


a'b' 


a  +  b  '  (a  +  bf  '      a{a-{-bf 
4,3  29  2 


(2a-\-b)b^x      Q       ,  bx 
^ '     \,^    =3ga7-j . 


x-8     2:r-16      24      3:^-24 


22.    5 


CHAPTER  X. 

Problems. 

Exercise    LXVII. 

Ex.  Find  the  number  the  sum  of  whose  third  and  fourth 
parts  is  equal  to  12. 

Let  X  =  the  number. 

Then  -  =  the  third  part  of  the  number, 

and  -  =  the  fourth  part  of  the  number, 

4 

.'.    -  +  -  =  the  sura  of  the  two  parts. 
3     4 


But 

12  = 

.-.    ?  +  ?  = 
3     4 

=  the  sum 

=  12. 

of  the  two 

parts, 

Multiply  both  sides  by  12 
4a  +  3a;  =144,    , 

7x  = 

=  144, 

.'.  x  = 

=  20f 

1.  Find  the  number  whose  third  and  fourth  parts  added 

together  make  14. 

2.  Find  the  number  whose  third  part  exceeds  its  fourth 

part  by  14. 

3.  The  half,  fourth,  and  fifth    of  a  certain  number  are 

together  equal  to  76  ;  find  the  number. 

4.  Find  the  number  whose  double  exceeds  its  half  by  12. 

5.  Divide  60  into  two  such  parts  that  a  seventh  of  one 

part  may  be  equal  to  an  eighth  of  the  other. 


138  ALGEBRA. 


6.  Divide  50  into  two  such  parts  tliat  a  fourth  of  one  part 

increased  by  five-sixths  of  the  other  part  may  be 
equal  to  40. 

7.  Divide  100  into  two  such  parts  that  a  fourth  of  one 

part  diminished  by  a  third  of  the  other  part  may  be 
equal  to  11. 

8.  The  sum  of  the  fourth,  fifth,  and  sixth  parts  of  a  cer- 

tain number  exceeds  the  half  of  the  number  by  112. 
What  is  the  number  ? 

9.  The  sum  of  two  numbers  is  5760,  and  their  difference  is 

equal  to  one-third  of  the  greater.  What  are  the 
numbers  ? 

10.  Divide   45   into    two    such   parts   that   the   first   part 

divided  by  2  shall  be  equal  to  the  second  part  mul- 
tiplied by  2. 

11.  Find  a  number  such  that  the  sum  of  its  fifth  and  its 

seventh  parts  shall  exceed  the  difference  of  its  fourth 
and  its  seventh  parts  by  99. 

12.  In  a  mixture  of  wine  and  water,  the  wine  was  25  gal- 

lons more  than  half  of  the  mixture,  and  the  water 

5  gallons  less  than  one-third  of  the  mixture.     How 

many  gallons  were  there  of  each  ? 

)k 

13.  In  a  certain  weight   of  gunpowder  the  saltpetre  was 

6  pounds  more  than  half  of  the  weight,  the  sulphur 
5  pounds  less  than  the  third,  and  the  charcoal  3 
pounds  less  than  the  fourth  of  the  weight.  How 
many  pounds  were  there  of  each  ? 

14.  Divide  46  into  two  parts  such  that   if  one    part   be 

divided  by  7,  and  the  other  by  3,  the  sum  of  the 
quotients  shall  be  10. 


PROBLEMS.  139 


15.  A  house  and  garden  cost  $  860,  and  five  times  the  price 

of  the  house  was  equal  to  twelve  times  the  price  of 
the  garden.     What  is  the  price  of  each  ? 

16.  A  man  leaves  the  half  of  his  property  to  his  wife,  a  sixth 

to  each  of  his  two  children,  a  twelfth  to  his  brother, 
and  the  remainder,  amounting  to  $600,  to  his  sister. 
What  w^as  the  amount  of  his  property  ? 

17.  The  sum  of  two  numbers  is  a  and  their  difference  is  h 

find  the  numbers. 

18.  Find  two  numbers  of  which  the  sum  is  70,  such  that 

the  first  divided  by  the  second  gives  2  as  a  quotient 
and  1  as  a  remainder. 

19.  Find  two  numbers  of  which  the  difference  is  25,  such 

that  the  second  divided  by  the  first  gives  4  as  a  quo- 
tient and  4  as  a  remainder. 

20.  Divide  the  number  208  into  two  parts  such  that  the 

sum  of  the  fourth  of  the  greater  and  the  third  of  the 
smaller  is  less  by  4  than  four  times  the  difference  of 
the  two  parts. 

21.  Find  four  consecutive  numbers  whose  sum  is  82. 

Note  I.  It  is  to  be  remembered  that  if  x  represent  a  person's  age 
at  the  present  time,  his  age  a  years  ago  will  be  represented  by  a;  —  a, 
and  a  years  hence  by  a;  +  a. 

Ex.  In  eight  years  a  boy  will  be  three  times  as  old  as  he 
was  eight  years  ago.     How  old  is  he  ? 
Let  X  =  the  number  of  years  of  his  age. 

Then  cc  —  8  =  the  number  of  years  of  his  age  eight  years  ago, 
and        x  +  8  =  the  number  of  years  of  his  age  eight  years  hence. 
.-.  a;  +  8    =  3  (a;  -  8), 
a; +  8    =3a;  +  24, 
a; -3a;  =  -24 -8, 
-2a;  =  -32, 
X  -  16. 


140  ALGEBRA. 


22.  A  is  72  years  old,  and  B's  age   is  two-thirds  of  A's. 

How  long  is  it  since  A  was  five  times  as  old  as  B  ? 

23.  A  mother  is  70  years  old,  her  daughter  is  half  that  age. 

How  long  is  it  since  the  mother  was  three  and  one- 
third  times  as  old  as  the  daughter  ? 

24.  A  father  is  three  times  as  old  as  the  son ;  four  years 

ago  the  father  was   four   times   as  old  as  the  son 
then  was.     What  is  the  age  of  each  ? 

25.  A  is  twice  as  old  as  B,  and  seven  years  ago  their  united 

ages  amounted  to  as  many  years  as  now  represent 
the  age  of  A.     Find  the  ages  of  A  and  B. 

26.  The  sum  of  the  ages  of  a  father  and  son  is  half  what  it 

will  be  in  25  years ;  the  difference  is  one-third  what 
the  sum  will  be  in  20  years.    What  is  the  age  of  each  ?  v7 

Note  II.  If  A  can  do  a  piece  of  work  in  x  days,  the  part  of  the 
work  that  he  can  do  in  one  day  will  be  represented  by  ^.  Thus,  if  he 
can  do  the  work  in  5  days,  in  1  day  he  can  do  \  of  the  work. 

Ex.  A  can  do  a  piece  of  work  in  5  days,  and  B  can  do  it 
in  4  days.  How  long  will  it  take  A  and  B  together 
to  do  the  work  ? 

Let       X  =  the  number  of  days  it  will  take  A  and  B  together. 
Then     ^  =  the  part  they  can  do  in  one  day. 
Now,    \  =  the  part  A  can  do  in  one  day, 
and  \  =  the  part  B  can  do  in  one  day. 

•'•   i  +  ¥  =  ^he  part  A  and  B  can  do  in  one  day. 

.••    i  +  i  =  i 
4a;  +  5x  =  20, 
9a;  =  20, 

x  =  2l 

Therefore  they  will  do  the  work  in  2-|  days. 

27.  A  can  do  a  piece  of  work  in  5  days,  B  in  6  days,  and 

C  in  7^  days ;  in  what  time  will  they  do  it,  all  work- 
ing together  ? 


PROBLEMS.  141 


28.  A  can  do  a  piece  of  work  in  2}  days,  B  in  3i  days,  and 

C  in  31  days  ;  in  what  time  will  they  do  it,  all  work- 
ing together  ? 

29.  Two  men  who  can  separately  do  a  piece  of  work  in  15 

days  and  16  days,  can,  with  the  help  of  another,  do 
it  in  6  days.  How  long  would  it  take  the  third  man 
to  do  it  alone  ? 

30.  A  can  do  half  as  much  work  as  B,  B  can  do  half  as 

much  as  C,  and  together  they  can  complete  a  piece 
of  work  in  24  days.  In  what  time  can  each  alone 
complete  the  work  ? 

31.  A  does  -f  of  a  piece  of  work  in  10  days,  when  B  comes 

to  help  him,  and  they  finish  the  work  in  3  days 
more.  How  long  would  it  have  taken  B  alone  to  do 
the  whole  work  ? 

32.  A  and  B  together  can  reap  a  field  in  12  hours,  A  and 

C  in  16  hours,  and  A  by  himself  in  20  hours.  In 
what  time  can  B  and  C  together  reap  it?  In  what 
time  can  A,  B,  and  C  together  reap  it  ? 

33.  A  and  B  together  can  do  a  piece  of  work  in  12  days, 

A  and  C  in  15  days,  B  and  C  in  20  days.  In  what 
time  can  they  do  it,  all  working  together  ? 

Note  III.  If  a  pipe  can  fill  a  vessel  in  a;  hours,  the  part  of  the 
vessel  filled  by  it  in  one  hour  will  be  represented  by  ^.  Thus,  if  a 
pipe  will  fill  a  vessel  in  3  hours,  in  1  hour  it  will  fill  ^  of  the  vessel. 

34.  A  tank  can  be  filled  by  two  pipes  in  24  minutes  and  30 

minutes  respectively,  and  emptied  by  a  third  in  20 
minutes.  In  what  time  will  it  be  filled  if  all  three 
are  running  together  ? 

35.  A  tank  can  be  filled  in  15.  minutes  by  two  pipes,  A  and 

B,  running  together.     After  A  has  been  running  by 


h 


142  ALGEBRA. 


itself  for  5  minutes,  B  is  also  turned  on,  and  the  tank 
is  filled  in  13  minutes  more.  In  what  time  may  it 
be  filled  by  each  pipe  separately  ? 

36.  A  cistern  could  be  filled  by  two  pipes  in  6  hours  and  8 

hours  respectively,  and  could  be  emptied  by  a  third 
in  12  hours.  In  what  time  would  the  cistern  be 
filled  if  the  pipes  were  all  running  together  ? 

37.  A  tank  can  be  filled  by  three  pipes  in  1  hour  and  20 

minutes,  3  hours  and  20  minutes,  and  5  hours,  re- 
spectively. In  what  time  will  the  tank  be  filled 
when  all  three  pipes  are  running  together  ? 

38.  If  three  pipes  can  fill  a  cistern  in  a,  b,  and  c  minutes, 

respectively,  in  what  time  will  it  be  filled  by  all 
three  running  together  ? 

39.  The  capacity  of  a  cistern  is  755^  gallons.     The  cistern 

has  three  pipes,  of  which  the  first  lets  in  12  gallons 
in  3 J  minutes,  the  second  15-J-  gallons  in  2-^  minutes, 
the  third  17  gallons  in  3  minutes.  In  what  time 
will  the  cistern  be  filled  by  the  three  pipes  running 
together  ? 

Note  IV.    In  questions  involving  distance,  time,  and  rate : 

Distance      m- 

— =  Time. 

Rate 

Thus,  if  a  man  travels  40  miles  at  the  rate  of  4  miles  an  hour, 
-—  =  number  of  ,h ours  required. 

Ex.  A  courier  who  goes  at  the  rate  of  31|-  miles  in  5  hours, 
is  followed,  after  8  hours,  by  another  who  goes  at  the 
rate  of  22-^  miles  in  3  hours.  In  how  many  hours 
will  the  second  overtake  the  first  ? 

Since  the  first  goes  31 1  miles  in  5  hours,  his  rate  per  hour  is  6^-q 
miles. 


PEOBLEMS.  143 


Since  the  second  goes  22}  miles  in  3  hours,  his  rate  per  hour  is  7} 
miles. 

Let  X  =  the  number  5f  hours  the  first  is  travelling. 

Then  a;  —  8  =  the  number  of  hours  the  second  is  travelling. 
Then  Q^jj  x  =  the  number  of  miles  the  first  travels  ; 
(x  —  8)  7}  =  the  number  of  miles  the  second  travels. 
They  both  travel  the  same  distance, 

.-.  6^^a;  =  (x-8)7}. 

The  solution  of  whicli  gives  42  hours. 

40.  A  sets  out  and  travels  at  the  rate  of  7  miles  in  5  hours. 

Eight  hours  afterwards  B  sets  out  from  the  same 
place  and  travels  in  the  same  direction,  at  the  rate 
of  5  miles  in  3  hours.  In  how  many  hours  will  B 
overtake  A  ? 

41.  A  person  walks  to  the  top  of  a  mountain  at  the  rate 

of  2J  miles  an  hour,  and  down  the  same  way  at  the 
rate  of  3^  miles  an  hour,  and  is  out  5  hours.  How 
far  is  it  to  the  top  of  the  mountain  ? 

42.  A  person  has  a  hours  at  his  disposal.     How  far  may  he 

ride  in  a  coach  which  travels  b  miles  an  hour,  so  as 
to  return  home  in  time,  walking  back  at  the  rate  of  c 
miles  an  hour? 

43.  The  distance  between  London  and  Edinburgh  is  360 

miles.  One  traveller  starts  from  Edinburgh  and 
travels  at  the  rate  of  10  miles  an  hour ;  another 
starts  at  the  same  time  from  London,  and  travels  at 
the  rate  of  8  miles  an  hour.  How  far  from  London 
will  they  meet  ? 

44.  Two  persons  set  out  from  the  same  place  in  opposite 

directions.  The  rate  of  one  of  them  per  hour  is  a 
mile  less  than  double  that  of  the  other,  and  in  4 
hours  they  are  32  miles  apart.  Determine  their 
rates. 


144 


ALGEBRA. 


45.  In  going  a  certain  distance,  a  train  travelling  35  miles 
an  hour  takes  2  hours  less  than  one  travelling  25 
miles  an  hour.     Determine  the  distance. 

Note  V.   In  problems  relating  to  clocks,  it  is  to  be  observed  that 
the  minute-hand  moves  twelve  times  as  fast  as  the  hour-hand. 

Ex.  Find  the  time  between  two  and  three  o'clock  when  the 
hands  of  a  clock  are  : 
I.    Together. 

II.   At  right  angles  to  each  other. 
III.    Opposite  to  each  other. 


Fig.  1. 


Fig.  2. 


Fig.  3. 


I.  Let  Cffand  CM  (Fig.  1)  denote  the  positions  of  the  hour  and 
minute  hands  at  2  o'clock,  and  CJB  the  position  of  both  hands  when 
together. 

Then  arc  IIB  =  one-twelfth  of  arc  MB. 

Let  X  =  number  of  minute-spaces  in  arc  3£B. 

Then  —  =  number  of  minute-spaces  in  arc  HB, 

and  10  =  number  of  minute-spaces  in  arc  MH. 

Now  arc  MB  =  arc  MH  -f  arc  HB. 


That  is, 


10 -f 


12r 


The  solution  of  this  equation  gives  x  =  10^^. 
Hence,  the  timel^  lOj"?"  ^linutes  past  2  o'clock. 

II.   Let  CB  and  CD  (Fig.  2)  denote  the  positions  of  the  hour  and 
minute  hands  when  at  right  angles  to  each  other. 


PROBLEMS.  145 


Let  X  =  number  of  minute-spaces  in  arc  MHBD. . 

Then  —  =  number  of  minute-spaces  in  arc  HB, 

and  10  ^  number  of  minute-spaces  in  arc  MIL 

15  =  number  of  minute-spaces  in  arc  BD. 
Now  arc  MHBD  =  arcs  MH  +  HB  +  BD. 
That  is,  a:  =  10  H-  A+  15. 

The  solution  of  this  equation  gives  x  =  27y\. 
Hence,  the  time  is  27-j^  minutes  past  2  o'clock. 

III.    Let  CB  and  CD  (Fig.  3)  denote  the  positions  of  the  hour  and 
minute  hands  when  opposite  to  each  other. 

Let  X  =  number  of  minute-spaces  in  arc  MHBD. 

Then  —  =  number  of  minute-spaces  in  arc  HB, 

and  10  =  number  of  minute-spaces  in  arc  MH, 

30  =  number  of  minute-spaces  in  arc  BD. 
Now  arc  MHBD  =  arcs  MH+  HB  +  BD. 
That  is,  a:  =  10-f--^'-t-30. 

The  solution  of  this  equation  gives  x  =  43^^. 
Hence,  the  time  is  43^  minutes  past  2  o'clock. 

46.  At  what  time  are  the  hands  of  a  watch  together : 

I.    Between  3  and  4  ? 

11.   Between  6  and  7? 

III.   Between  9  and  10? 

47.  At  what  time  are  the  hands  of  a  watch  at  right  angles : 

I.    Between  3  and  4  ? 

II.    Between  4  and  5  ? 

III.    Between  7  and  8? 

48.  At  what  time  are  the  hands  of  a  watch  opposite  to 

each  other :  0 

I.  Between  1  and  2  ? 

II.  Between  4  and  5  ? 

III.  Between  8  and  9? 


146  ALGEBRA. 


49.  It  is  between  2  and  3  o'clock ;  but  a  person  looking  at 

bis  watch  and  mistaking  the  hour-hand  for  the 
minute  hand,  fancies  that  the  time  of  day  is  55 
minutes  earlier  than  it  really  is.  What  is  the  true 
time? 

Note  VI.  It  is  to  be  observed  that  if  a  represent  the  number  of 
feet  in  the  length  of  a  step  or  leap,  and  x  the  number  of  steps  or  leaps 
taken,  then  ax  will  represent  the  number  of  feet  in  the  distance 
made. 

Ex.  A  hare  takes  4  leaps  to  a  greyhound's  3 ;  but  2  of  the 
greyhound's  leaps  are  equivalent  to  3  of  the  hare's. 
The  hare  has  a  start  of  50  leaps.  How  many  leaps 
must  the  greyhound  take  to  catch  the  hare  ? 

Let  3  a;  =  the  number  of  leaps  taken  by  the  greyhound. 
Then  4  a;  =  the  number  of  leaps  of  the  hare  in  the  same  time. 
Also,  let  a  denote  the  number  of  feet  in  one  leap  of  the  hare. 

Then  —  will  denote  the  number  of  feet  in  one  leap  of  the  grey- 
hound. 

That  is,  3  a;  X  -^  =  the  whole  distance, 

and  (50  +  4.a;)  a  =  the  whole  distance, 

...  ^-^  =  (bO-v^x)a. 

Divide  by  a,  —  =  50  + 4  a?, 

A 

9a;=100  +  8a;, 
a;=100, 
.-.  3x  =  300. 

Thus  the  greyhound  must  take  300  leaps. 

50.  A  hare  takes  6  leaps  to  a  dog's  5,  and  7  of  the  dog's 

leaps  are  equivalent  to  9  of  the  hare's.  The  hare 
has  a  start  of  50  of  her  own  leaps.  How  many  leaps 
will  the  hare  take  before  she  is  caught  ? 


PROBLEMS.  147 


51.  A  greyhound  makes  3  leaps  while  a  hare  makes  4  ;  but 

2  of  the  greyhound's  leaps  are  equivalent  to  3  of  the 
hare's.  The  hare  has  a  start  of  50  of  the  greyhound's 
leaps.  How  many  leaps  does  each  take  before  the 
hare  is  caught  ? 

52.  A  greyhound  makes  two  leaps  while  a  hare  makes  3 ; 

but  1  leap  of  the  greyhound  is  equivalent  to  2  of  the 
hare's.  The  hare  has  a  start  of  80  of  her  own  leaps. 
How  many  leaps  will  the  hare  take  before  she  is 
caught  ? 

Note  VII.  It  is  to  be  observed  that  if  the  number  of  units  in  the 
breadth  and  length  of  a  rectangle  be  represented  by  x  and  x  +  a, 
respectively,  then  x{x  ■\-  a)  will  represent  the  number  of  surface  units 
in  the  rectangle,  the  unit  of  surface  having  the  same  name  as  the 
linear  unit  in  which  the  sides  of  the  rectangle  are  expressed. 

53.  A  rectangle  whose  length  is  5  feet  more  than  its  breadth 

would  have  its  area  increased  by  22  feet  if  its  length 
and  breadth  were  each  made  a  foot  more.  Find  its 
dimensions. 

54.  A  rectangle  has  its  length  and  breadth  respectively  5 

feet  longer  and  3  feet  shorter  than  the  side  of  the 
equivalent  square.     Find  its  area. 

55.  The  length  of  a  rectangle  is  an  inch  less  than  double 

its  breadth ;  and  when  a  strip  3  inches  wide  i!s  cut 
off  all  round,  the  area  is  diminished  by  210  inches. 
Find  the  size  of  the  rectangle  at  first. 

56.  The  length  of  a  floor  exceeds  the  breadth  by  4  feet ;  if 

each  dimension  were  increased  by  1  foot,  the  area 
of  the  room  would  be  increased  by  27  square  feet. 
Find  its  dimensions. 

Note  VIII.  It  is  to  be  observed  that  if  h  pounds  of  metal  lose  a 
pounds  when  weighed  in  water,  1  pound  will  lose  ^  of  a  pounds, 
or  5  of  a  pound. 


148  ALGEBRA, 


57.  A  mass  of  tin  and  lead  weighing  180  pounds  loses  21 

pounds  when  weighed  in  water ;  and  it  is  known  that 
37  pounds  of  tin  lose  5  pounds,  and  23  pounds  of 
lead  lose  2  pounds,  when  weighed  in  water.  How 
many  pounds  of  tin  and  of  lead  in  the  mass  ? 

58.  If  19  pounds  of  gold  lose  1  pound,  and  10  pounds  of 

silver  lose  1  pound,  when  weighed  in  water,  find  the 
amount  of  each  in  a  mass  of  gold  and  silver  weighing 
106  pounds  in  air  and  99  pounds  in  water. 

59.  Fifteen  sovereigns  should  weigh  77  pennyweights ;  but 

a  parcel  of  light  sovereigns,  having  been  weighed  and 
counted,  was  found  to  contain  9  more  than  was  sup- 
posed from  the  weight ;  and  it  appeared  that  21  of 
these  coins  weighed  the  same  as  20  true  sovereigns. 
How  many  were  there  altogether  ? 


60.  There  are  two  silver  cups,  and  one  cover  for  both.    The 

first  weighs  12  ounces,  and  with  the  cover  weighs 
twice  as  much  as  the  other  without  it ;  but  the  sec- 
ond with  the  cover  weighs  one-third  more  than  the 
first  without  it.     Find  the  weight  of  the  cover. 

61.  A  man  wishes  to  enclose  a  circular  piece  of  ground  with 

palisades,  and  finds  that  if  he  sets  them  a  foot  apart 
he  will  have  too  few  by  150 ;  but  if  he  sets  them  a 
yard  apart  he  will  have  too  many  by  70.  What  is  the 
circuit  of  the  piece  of  ground  ? 

62.  A  horse  was  sold  at  a  loss  for  $  200 ;  but  if  it  had  been 

sold  for  $250,  the  gain  would  have  been  three-fourths 
of  the  loss  when  sold  for  $200.  Find  the  value  of 
the  horse. 

63.  A  and  B  shoot  by  turns  at  a  target.     A  puts  7  bullets 

out  of  12,  and  B  9  out  of  12,  into  the  centre.  Be- 
tween them  they  put  in  32  bullets.  How  many  shots 
did  each  fire  ? 


PROBLEMS.  149 


64.  A  boy  buys  a  number  of  apples  at  the  rate  of  5  for  2 

pence.  He  sells  half  of  them  at  2  a  penny  and  the 
rest  at  3  a  penny,  and  clears  a  penny  by  the  tran- 
saction.    How  many  does  he  buy  ? 

65.  A  person  bought  a  piece  of  land  for  ?  6750,  of  which 

he  kept  f  for  himself.  At  the  cost  of  $  250  he  made 
a  road  which  took  y^^  of  the  remainder,  and  then  sold 
the  rest  at  12-^  cents  a  square  yard  more  than  double 
the  price  it  cost  him,  thus  clearing  his  outlay  and 
$500  besides.  How  much  land  did  he  buy,  and 
what  was  the  cost-price  per  yard  ? 

66.  A  boy  who  runs  at  the  rate  of  12  yards  per  second 

starts  20  yards  behind  another  whose  rate  is  10^ 
yards  per  second.  How  soon  will  the  first  boy  be 
10  yards  ahead  of  the  second  ? 

67.  A  merchant  adds  yearly  to  his  capital  one-third  of  it, 

but  takes  from  it,  at  the  end  of  each  year,  $5000  for 
expenses.  At  the  end  of  the  third  year,  after  de- 
ducting the  last  $5000,  he  has  twice  his  original 
capital.     How  much  had  he  at  first  ? 

68.  A  shepherd  lost  a  number  of  sheep  equal  to  one-fourth 

of  his  flock  and  one-fourth  of  a  sheep  ;  then,  he  lost  a 
number  equal  to  one-third  of  what  he  had  left  and 
one-third  of  a  sheep  ;  finally,  he  lost  a  number  equal 
,  to  one-half  of  what  now  remained  and  one-half  a 
sheep,  after  which  he  had  but  25  sheep  left.  How 
many  had  he  at  first? 

69.  A  trader  maintained  himself  for  three  years  at  an  ex- 

pense of  $  250  a  year ;  and  each  year  increased  that 
part  of  his  stock  which  was  not  so  expended  by  one- 
third  of  it.  At  the  end  of  the  third  year  his  original 
stock  was  doubled.     What  was  his  original  stock  ? 


150  ALGEBRA. 


70.  A  cask  contains  12  gallons  of  wine  and  18  gallons  of 

water ;  another  cask  contains  9  gallons  of  wine  and 
3  gallons  of  water.  How  many  gallons  must  be 
drawn  from  each  cask  to  produce  a  mixture  contain- 
ing 7  gallons  of  wine  and  7  gallons  of  water  ? 

71.  The  members  of  a  club  subscribe  each  as  many  dollars 

as  there  are  members.  If  there  had  been  12  more 
members,  the  subscription  from  each  would  have 
been  $10  less,  to  amount  to  the  same  sum.  How 
many  members  were  there  ? 

72.  A  number  of  troops  being  formed  into  a  solid  square, 

it  was  found  there  were  60  men  over;  but  when 
formed  in  a  column  with  6  men  more  in  front  than 
before,  and  3  men  less  in  depth,  there  was  lacking 
one  man  to  complete  it.     Find  the  number  of  troops. 

73.  An  officer  can  form  the  men  of  his  regiment  into  a  hol- 

low square  twelve  deep.  The  number  of  men  in  the 
regiment  is  1296.  Find  the  number  of  men  in  the 
front  of  the  hollow  square. 

74.  A  person  starts  from  P  and  walks  towards  Q  at  the 

rate  of  3  miles  an  hour;  20  minutes  later  another 
person  starts  from  Q  and  walks  towards  P  at  the 
rate  of  4  miles  an  hour.  The  distance  from  P  to  Q, 
is  20  miles.     How  far  from  P  will  they  meet  ? 

75.  A  person  engaged  to  work  a  days  on  these  conditions : 

for  each  day  he  worked  he  was  to  receive  h  cents, 
and  for  each  day  he  was  idle  he  was  to  forfeit  c  cents. 
At  the  end  of  a  days  he  received  d  cents.  How 
many  days  was  he  idle  ? 

76.  A  banker  has  two  kinds  of  coins :  it  takes  a  pieces  of 

the  first  to  make  a  dollar,  and  h  pieces  of  the  second 
to  make  a  dollar.  A  person  wishes  to  obtain  c  pieces 
for  a  dollar.  How  many  pieces  of  each  kind  must 
the  banker  give  him  ? 


CHAPTER  XI. 
Simultaneous  Equations  of  the  First  Degree. 

184.  If  one  equation  contain  two  unknown  quantities,  an 
indefinite  number  of  pairs  of  values  may  be  found  that  will 
satisfy  the  equation. 

Thus,  in  the  equation  x  -\-  y  =  \0,  any  values  may  be 
given  to  x,  and  corresponding  values  for  y  may  be  found. 
Any  pair  of  these  values  substituted  for  x  and  y  will  satisfy 
the  equation. 

185.  But  if  a  second  equation  be  given,  expressing  differ- 
ent relations  between  the  unknown  quantities,  only  one  pair 
of  values  of  x  and  y  can  be  found  that  will  satisfy  both 
equations. 

Thus,  if  besides  the  equation  a:  -f-  y  =  10,  another  equa- 
tion, x  —  y=^2,  be  given,  it  is  evident  that  the  values  of 
X  and  y  which  will  satisfy  both  equations  are 

x  =  %\ 

for  6  +  4  =  10,  and  6  —  4  =  2;  and  these  are  the  only  val- 
ues of  X  and  y  that  will  satisfy  both  equations. 

186.  Equations  that  express  different  relations  between 
the  unknown  quantities  are  called  independent  equations. 

Thus,  rr  -f  y  =  10  and  a;  —  y  =  2  are  independent  equa- 
tions ;  they  express  different  relations  between  x  and  y. 
But  X  -\-  y=\^  and  3:r  -f  3y  =  30  are  not  independent 


152  ALGEBEA. 


equations;  one  is  derived  immediately  from  the  other, 
and  both  express  the  same  relation  between  the  unknown 
quantities. 

187.  Equations  that  are  to  be  satisfied  by  the  savie  val- 
ues of  the  unknown  quantities  are  called  simultaneous 
equations. 

188.  Simultaneous  equations  are  solved  by  combining 
the  equations  so  as  to  obtain  a  single  equation  containing 
only  one  unknown  quantity ;  and  this  process  is  called 
elimination. 

Three  methods  of  elimination  are  generally  given  : 
I.    By  Addition  or  Subtraction. 
II.   By  Substitution. 
III.    By  Comparison. 

Elimination  by  Addition  or  Subtraction. 
(1)    Solve-, 


2a: 

-3y  = 

=    4T 
=^32/ 

(1) 

Zx 

+  2y^ 

(2) 

Multiply  (1)  by  i 

I  and  (2)  by  3, 

4a:-6y  = 

:            8 

(3) 

9rK  +  6y  = 

:        96 

(4) 

Add  (3)  and  (4), 

13  a; 

104 

Substitute  the  value  of  x  in  (2), 

24 +  23/ =  32, 
•■•  3/ =  4. 
In  this  solution  y  is  eliminated  by  addition. 

(2)    Solve:  6:^  +  SSy  =  177  j  (1) 

8:r-21y=    33/  (2) 

Multiply  (1)  by  4  and  (2)  by  3, 

24i  + 1402/ =  708  (3) 

24a;-    63y=    99  (4) 

Subtract  (4)  from  (3),  203  y  =  609 

•••  2/ =  3. 


SIMULTANEOUS   EQUATIONS.  153 

Substitute  the  value  oft/  in  (2). 
8a;- 63  =  33. 
.-.  X  =  12. 
In  this  solution  x  is  eliminated  by  subtraction. 

189.  Hence,  to  eliminate  an  unknown  quantity  by  addi- 
tion or  subtraction, 

Multiply  the  equations  by  such  numbers  as  will  mahe  the 
coefficients  of  this  unknown  quantity  equal  in  the  resulting 
equations. 

Add  the  resulting  equations,  or  subtract  one  from  the  other, 
according  as  these  equAxl  quantities  have  unlike  or  like  signs. 

Note.  It  is  generally  best  to  select  that  unknown  quantity  to  be 
eliminated  which  requires  the  smallest  multipliers  to  make  its  coeffi- 
cients equal ;  and  the  smallest  multiplier  for  each  equation  is  found 
by  dividing  the  L.  C.  M.  of  the  coefficients  of  this  unknown  quantity 
by  the  given  coefficient  in  that  equation.  Thus,  in  example  (2),  the 
L.  C.  M.  of  6  and  8  (the  coefficients  of  x),  is  24,  and  hence  the  smallest 
multipliers  of  the  two  equations  are  4  and  3  respectively. 

Sometimes  the  solution  is  simplified  by  first  adding  the 
given  equations,  or  by  subtracting  one  from  the  other. 

(3) 


) 

a;  +  49y=    51 
40.T+      y=   99 

(1) 
(2) 

Add  (1)  and  (2), 
Divide  (3)  by  50, 
Subtract  (4)  from  (1), 

Subtract  (4)  from  (2), 

50a;  +  50y  =  150 
x  +  y  =  3. 
48y  =  48, 

•••  y  =  i. 

48  a;  =  96, 
.-.  x  =  2. 

(3) 
(4) 

Exercise  LXVIII. 
Solve  by  addition  or  subtraction  : 
1.  2xi-Sy=-7)     3.    7a;  +  2?/  =  30|    5.   5a:  +  4y  =  58| 
J  v-Sx=   2)  3a:  +  77/=67J 


4:X-5y  =  S)  y-3x=   2J  Sx  +  7y  = 

x  —  2y  =  4,)     4.    3:r  — 5y  =  51|    6.    3:r 
2x-    y  =  5J  2x+7y=   3J  3y 


154  ALGEBBA. 


7. 

3a;-4y  =  -51 
4a;-5y=      li 

11. 

\2x-\-    7y  =  176-) 
3y-19ru-      3/ 

8. 

lla;  +  32/-100| 
^x—1y=     4J 

12. 

2x-ny^    %\ 
4y-9:r  =  19i 

9. 

^  +  49y  =  693| 
49a;+      y  =  357J 

13. 

69y-17a;=:    103  | 
14a;-13y  =  -4lJ 

10. 

16y-3a;  =  23J 

14. 

17a;  +  30y  =  591 
19a;  +  28y=77J      ^ 

Elimination  by  Substitution. 

(1)    Solve:  2^  +  3y  =  8| 

3a;4-7y=7  J 

2rr  +  32/  =  8  (1) 

3aj  +  72/  =  7  (2) 

Transpose  3  y  in  (1),  2  a;  =  8  —  3  y. 

o  o  . . 

Divide  by  coefficient  of  x,      x=    ~    •'.  (4) 


Substitute  the  value  of  x 


m(2).3(^^)+7y 


24 


^  +  72/  =  7. 


2 

24-9y  +  142/  =  14, 
52/  = -10, 
•••  2/--2. 
Substitute  the  value  of  y  in  (1),  2»  —  6  =  8, 

.-.  x=7. 

190.  Hence,  to  eliminate  an  unknown  quantity  by  sub- 
stitution, 

I'rom  one  of  the  equations  obtain  the  value  of  one  of  the 
unknown  quantities  in  terms  of  the  other. 

Substitute  for  this  unknown  quantity  its  value  in  the  other 
equation,  and  reduce  the  resulting  equation. 


simultaneous  equations.  155 

Exercise  LXIX. 
Solve  by  substitution : 

1.  3x-iy  =  2)  8.    3a: -4?/ ^18  I 

2.  7a: -52/ =  24  I  9.    9a:-5y=-52| 
4a:-3y  =  llJ  8y-3a:=    8J 

3.  3a:  +  2y  =  32"l  10.      5x-dy=   4| 
20a: -3y=    IJ  12y-7a:  =  10J 

4.  lla:-7y  =  37|  11.      9y-7a:  =  131 

8a:  +  9y  =  41J  15a;-7y=    9J 

5.  7a:+    5y  =  60'l  12.      5a:-2y=    511 
13a:-lly  =  10/  19a:-3y  =  180J 

6.  6a:-7y  =  42)  13.   4a:+    9^=1061 
7a:  -  6y  =  75  J  8a:  +  17y  =  198  J 

7.  10a:+    9y  =  2901  14.      8a:  +  3y  =  31 
12a:  -  lly  =  130  J  12a:  +  9y  =  3  J 


Elimination  by  Comparison. 

9y 
4y 


Solve:  2a:-9y  =  ll| 

Sx-4:V=7    ) 


2x-9y  =  n,  (1) 

3a:-4y  =  7.  (2) 

Transpose  9  3/  in  (1)  and  42/  in  (2),  2  a:  =  11  +  9y,  (3) 

3x  =  7  +  43/.  (4) 

Divide  (3)  by  2  and  (4)  by  3,  x  =  ^^^^^,  (5) 

^=LLiy.  (6) 

Equate  the  values  of  x.  "  +  ^'V  =  '^■^^.  0) 


156  ALGEBRA. 


Reduce  (7)  33  +  27y  =  U  +  8  ij, 

19y  =  -19, 

•••  y--i. 

Substitute  the  value  of  y  in  (1),  2  a;  +  9  =  11, 

.-.  x=l. 

191.  Hence,  to  eliminate  an  unknown  quantity  by  com- 
parison, 

Froin  each  equation  obtain  the  value  of  one  of  the  unknown 
quantities  in  terms  of  the  other. 

Form  an  equation  from,  these  equal  values  and  reduce  the 
equation. 

Note.   If,  in  the  last  example,  (3)  be  divided  by  (4),  the  resulting 

equation,  -  = ^,  would,  when  reduced,  eive  the  value  of  y. 

^  '3       7+43/  '  ^  ^ 

This  is  the  shortest  method,  and  therefore  to  be  preferred. 

Exercise  LXX. 
Solve  by  comparison : 

1.  a:  +  15y  =  53|  8.    2>y-lx=    41 

2.  4^+    9y  =  51|  9.   21y +  20;r  =  165  | 
807-133/=    9/  77y-30a;  =  295j 

3.  4a;  +  3y  =  481  10.    Ilx-I0y  =  l^\ 
by-3x  =  22}  bx+    1^  =  4:1) 

4.  2x  +  Sy  =  iS\  11.    7y-3a:  =  1391 
10a;-    y=    7J  2x-i-^2/=    91 J 

5.  5x-    7y=   831  12.    17:z;+12y=    59) 
lla;+12y  =  100/  19^-   4y  =  153J 

6.  5x+7y  =  AS\  13.   24:r+    7y=   271 
J  8:r-33v  =  115i 


lla;4-9y  =  69J  Sx-SSy 

8a;-21y=    33  \  14.   a;  =  3y-19 

6:r  +  35y  =  177J  y==3a;-23 


SIMULTANEOUS   EQUATIONS.  157 

192.    Each  equation  must  be  simplified,  if  necessary,  be- 
fore the  elimination  is  performed. 

(1)    Solve:   (a;-l)(y  +  2)  =  (:r-3)(y-l)  +  8 

5  4 

(a:-l)(y  +  2)  =  (x-3)(y-l)  +  8  (1) 

2a:-l_3(y-2)^^  (2) 

5  4 

Simplify  (1),  xy  +  2x-y -2  =  xy -x -3y  +  3  +S. 

Transpose  and  combine,  3a;  +  2y  =  13.  (3) 

Simplify  (2),  8a;-4- ISy  +  30  =  20. 

Transpose  and  combine,  8a;  —  15y  =  —  6.  (4) 

Multiply  (3)  by  8,  24  a;  +  16  y  =  104.  (5) 

Multiply  (4)  by  3,  2^x-ioy  =  -18.  (6) 

Subtract  (G)  from  (5),  61  y  =  122, 

.-.  y  =  2. 

Substitute  the  value  of  y  in  (3),   3  x  +  4  =  13, 

.',  x  =  3. 

Exercise  LXXI. 
Solve : 

1.     X(7J  +  7)  =  1/{X+1) 


2a;  +  20  =  3y  +  l      i  IT     o^ 


5(a:+3)  =  3.(y-2)  +  2J. 

2.    2x-^-^=0)  4.   ^-^  =  0 

5  I  5  10 

3y+^_9=oJ  1  +  2^=2^3 


5.   (a:  +  l)(y  +  2)-(:.  +  2)(y-M)=-n 
3(a;  +  3)-4(y  +  4)  =  -8  J 

x-2  10-a:^?/-10 

'       5  3              4 

2?/  +  4  2x~{-7/^x-\-lS 

3  8              4 


158 


ALGEBRA. 


3  4  5 
x  —  2>     y  —  3 

4  ~     3 


27J-X 


15    ^-4__y  +  2 
5  10 


o    3rc  — 2v  ,  5a7  — 3y  ,  t   ^ 

8.   -r-^-l ^^  =  x-\-l 

5  3 

2x  —  '^y  ,  4:r  — 3v  ,  T 

~3       +— ^   =^+^ 


16.   3^±12^^9        ^ 

l-3a;_^ll-3y 
7  5 


3  4 

3a;-4y4-3  ,  4a;-2y-9_. 
4  "^  3 


10.    li^  =  liy  +  4A| 
^x  =  \y-2l-J^i 


17.   5^-i(5y  +  2)  =  321 
3y  +  i(:.  +  2)=9      / 


11. 


13 


x-\-2y-\-2>  4a:-5y  +  6 

3  _  19 

6a;-5y  +  4      3a;  +  2y+l 


18.    3  a;— .25  y  =  28 


.12:r+.7y^ 


.54/ 


12.   ^±^  =  1^ 

y  —  X      8 

9^_3y_±i4^ioo 


19.    7(^-l)  =  3(y  +  8) 

4^  +  2^5y+9 

9  2 


3:r-5y      o_2^  +  y  y 
^^-         2       ^  5 

Q     ^-2y^.r  ,  y 
4  2^3 


20.    7:r  +  i(2y+4)-16 


3y 


K2y+4)-16| 
i(a;  +  2)=8     J 


4a;-3y-7_3x      2y_5 
5  10      15      6 

3     ^2     20  15    ^6^10 


SIMULTANEOUS   EQUATIONS. 


159 


21. 


22. 


507  —  6 


2y-2 
3a; -19      .      3.y- 


13 

5a:  +  6y      3a7  — 2y 
6  4 


5rr 


4- 


2  2  3 

2x-\-y     9a:-7_3(y  +  3)      4a:  +  5y 
2    "  8  4  16 

23.   3y  +  ll=^^-y(^  +  ^y)  +  31-4rr 
^  a;-y  +  4 

(:r+7)(y-2)  +  3-2:ry-(y-l)(:r+l) 


^^    6.r  +  9  J  3a:  +  5?/_3^  j  3:r  +  4 
4           4a:  — 6  2 

8y-f7  ,  6rg-3y_,  4y-9 
10     "^  2y-8  5 


25. 


20 


59 -2a; 


y+ 


23 -a:  z 

y-3  _3Q      73-3y 
^-18  3 


Literal  Simultaneous  Equations. 

193.    The  method  of  solving  literal  simultaneous  equa 
tions  is  as  follows : 

Solve :  ax-{-'by^=7n\ 

cx-{-dy  =  n  ) 

ax  +  by  =  m 

ex  +  dy  =  n 
Multiply  (1)  by  c,  acx  +  bey  =  cm 

Multiply  (2)  by  a,  acx  +  ady  =  an 

Subtract  (4)  from  (3), 


(1) 
(2) 

(3) 
(4) 


{he  —  ad)  y  =  cm  —  an 


Divide  by  coefficient  of  y, 


y  = 


cm  ~  an 
be  — ad 


160 


ALGEBRA. 


To  find  the  value  of  x : 

Multiply  (1)  by  d, 
Multiply  (2)  by  &, 
Subtract  (6)  from  (5), 

Divide  by  coefficient  of  x. 


adx  +  hdy  =  dm 
hex  +  hdy  =  bn 
{ad  —bc)x  =  dm  —  bn 
_  dm  —  bn 
ad—  be 


(5) 
(6) 


Solve : 

1.   x-j~^  =  a 
X  —  y  "^h 

%    ax  -\-hy  =  c 
px-^qy  =  r 


a      0 


Exercise  LXXII. 

3.   rax  +  ny  =  a  I     5.   rax  —  ny=^r    \ 
px-\-  qy  =  h    }  ra'x  +  n\/  =  r' } 


4.   ax-{-by  =  e\       6.    ax-i'by  =  c 


ax-i-cy  =  d 


dx  +fy  =  c- 


b  '  a 


--    x  —  y-^1 

12.    ^   '   ^  =a^ 

x-y-1  { 

x  +  y  —  l 


8.    abx  -{-  cdy  =  2 

d-b 

ax  —  cy^= 

^         bd 


b-\-y      Za-\-x 
ax-{-2by^=d 


13.    ax^=by  -\- 


a^-\-b'' 


{a  —  b)  X  =^  {a -{- b)  y 

14.    ax-[~by  =^  c^ 
a  b 


b-\-y      a-\-x 


10. 


^=JL 


a-{-b      a  —  b      a-\-b 

^     ■     y    -    1 


15. 


a+^      a—b 

x  —  y 37-f-y 

2a^)  ~a2_|.^2 


-^  =  2^1 


a  +  6      a  —  6      a  —  b 

11.   a(a— a;)  =  5(a;+3/— a)  )   16.    bx  —  bc=^ay  —  ac') 
a(y — b—x)=b(y—b)  J  x  —  y  =  a  —  b         J 


SIMULTANEOUS   EQUATIONS.  161 


17.    ^-« 


y-h 

a  {x  —  oi)  -\rh  {;\j  ~h)  ■\-  abc  =  0 

la    {a-\-h)x  —  {a  —  h)y  =  4:ab  ) 

{a-b)x  +  {a  +  h)y  =  2o?~2h^) 

19.  {x-\-a){y-\-h)-{x-a){i/-h)  =  2{a-hf}^ 
:r-3/+2(a-5)=0  / 

20.  {a-\-h){x  +  y)-{a-b){x-y)  =  a'\   ' 
{a-b){x-\-y)  +  {a  +  b){x-y)  =  lP  I 

194.  Fractional  simultaneous  equations,  of  which  the  de- 
nominators are  simple  expressions  and  contain  the  unknown 
quantities,  may  be  solved  as  follows : 

(1)   Solve:  -  +  -  =  m 

X     y 

c   ,  d 
--f-  =  n 
X     y 

-  +  -  =  m.  (1) 


y 


-  +  -  =  ^-      •  (2) 


X 


y 


Multiply  (1)  by  c,  '^^'^Jl^f^,  (3) 

X       y 

Multiply  (2)  by  a,  ^  +  ^  =  an.  (4) 

X       y 

Subtract  (4)  from  (3),  — — =  cm  —  an. 

y     . 

Multiply  both  sides  by  y,     bc  —  ad  =  {cm  —  an)  y, 

be  —  ad 

.•.  y  = . 

cm  —  an, 

Multiply  (1)  by  (f,  ^A  +  M=dm.  (5) 

X        y 

Multiply  (2)  by  6,  ^  +  ^  =  bn.  (6) 

X       y 


162 


ALGEBRA. 


Subtract  (G)  from  (5), 


ad—  he 


=  dm  —  hn. 


Multiply  both  sides  by  x,     ad—hc^  {dm  —  hn) x, 

ad— he 
.•.X 


dm  —  bn 


(2)     Solve 


3^7      5y 


_7_ 
6x 


Multiply  (2)  by  4, 
Add  (1)  and  (3), 
Divide  both  sides  by  19, 


1 

lOy 


6x      by 


3a;  by 
19 
3a; 
jL 
3a; 


Substitute  the  value  of  x  in  (1), 

Transpose, 

Divide  both  sides  by  2, 


19. 
1, 

by 
by 


Solve : 

1.  1  +  ?  =  10 

X      y^ 

X      y 

2.  !  +  ?  =  « 

X      y 
X     y 


Exercise  LXXIII. 


3.  ?_A 
1 


a; 


4^      y 
4.1  +  2  =  4 

X      y 


27 

n 

72  J 


5.    ?-^  =  5- 

^    y 

1-5  =  6 

6.      _-|-_=r_ 

X  y  0 
h  a  _^hc 
X     y       a 


(1) 
(2) 
(3) 


SIMULTANEOUS  EQUATIONS. 


163 


7.     —  +-_=:5 

ax     by 
ax      hy 


8. j =7n-\-n 

nx      my 

n\  m         2  I     2 
X      y 


m 


9.  ^  +  ^ 

X      y 

h      a  _ 

X     y~ 


195.  If  three  simultaneous  equations  are  given,  involv- 
ing three  unknown  quantities,  one  of  the  unknown  quanti- 
ties must  be  eliminated  between  two  pairs  of  the  equations ; 
then  a  second  between  the  resulting  equations. 

196.  Likewise,  if  four  or  more  equations  are  given, 
involving  four  or  more  unknown  quantities,  one  of  the 
unknown  quantities  must  be  eliminated  between  three  or 
more  pairs  of  the  equations ;  then  a  second  between  the 
pairs  that  can  be  found  of  the  resulting  equations ;  and 
so  on. 


Solve:                    2  a:- 
3a;- 
5a:- 

-3?/  +  4z=:   4] 
f  5y-7z  =  12  [ 
_    y--8z=    6) 

(1) 

(2) 
(3) 

Eliminate  z  between  two 

pairs  of  these  equations. 

Multiply  (1)  by  2, 

(3)  is 

Add, 

4a;-6y  +  82=    8 
5x-    .7-82=    5 
9x-7y          =13 

(4) 
(5) 

Multiply  (1)  by  7, 
Multiply  (2)  by  4, 
Add, 

14x-21y  +  282  =  28 
12a; +  20.7-282  =  48 
26a;-      y            =76 

(6) 

Multiply  (6)  by  7, 

(5)  is 

Subtract  (5)  from  (7), 

182x-7y  =  532 

9x-7v=    13. 

173x          =  519 

(7) 

Substitute  the  value  of  x  in  (6),                    78  —  y  = 

=  76, 

•••  y  = 

Substitute  the  values  of  x  and  y  in  (1),  6  —  6  +  43  = 

.•.  z  = 

=  4, 
=  1. 

164  ALGEBRA. 


c.  I     .  Exercise  LXXIV. 

1.  5a:  +  3y-6z  =  4  1  10.    3a;-y  +  z-=17 

a;-2y  +  2z  =  2     J  7a;  +  4y-50  =  3 

2.  4a;  — 5y  +  2z  =  6     ^  11.   a7  +  y  +  z  =  5 
2a:-f3y-0=2O     I  3a;-5y+7z=75 


7a;-4y  +  3z  =  35  J  9a;-ll2  +  10  =  0 

3.    5;  +  y  +  z  =  6  ^         12.   rr  +  2y  +  30  =  6 

5:i:  +  4y  +  30  =  22       I  2a:  +  4y  +  2z  =  8 


15^+10y  +  62  =  53  J  3a;  +  2y +  82  =  101  J 

4.  4:t'  — 3?/  +  z  =  9     1  13.   ru  — 33/  — 22  =  1 

9:r  +  y-52  =  16   I  2^;- 3y  +  5^  =  - 19 

a:-43/  +  32-=2     J  5:r  +  2y-2  =  12 

5.  8x  +  4:y-3z^Q  ^  14.  Sx-27/  =  6  ^ 
^  +  3y-2  =  7  I  4a;-3?/  +  22;  =  ll  I 
4^'-5y  +  4z  =  8j                     x-27/-bz  =  —  1  j 

6.  12a;  +  5y-40  =  29  ^  15.  x  +  y=l 
13a:-2y  +  52  =  58  I  y  +  z^=9 
17a:  — 3/  — 2  =  15        J  a;  +  z=5 

7.  y  — a;  +  z  =  — 5     -j  16.    2a:  — 3^  =  3 
z  — y  — 0:  =  - 25  I  3y  — 42=7 


x-\-y  +  z  =  36      )  4:z-5x  =  2 

a   a:  +  y  +  2  =  30  ^  17.    3a:-4y+62  =  l 

8a:  +  4y  +  2z  =  50     [  2a:  +  2y-2  =  l 

.     27a:  +  9y  +  32;  =  64  J  7a:-6y+72  =  2 

9.    15y  =  242-10a:  +  41  ^     18.   7a:-3y  =  30 

15a:  =  12y-162  +  10  I  9y-52  =  34 

18a:-(72-13)  =  14y  J  a:  +  y  +  z  =  33 


SIMULTANEOUS   EQUATIONS. 


165 


19.    a:  +  |  +  |  =  6 

23. 

^-f  +  l  =  7t 

>! 

2      3 

a;      5y  '  z 

^+l+I=-i 

1          1        ^ 
3-.  +  ^y  +  .^  =  ^°* 

,+|+|'=n 

^_1+4  =  16,V 

2     3 

5a;      2y  '  z           ^" 

■ 

20.   1  +  2  =  5      1 

24. 

2_3  +  l  =  2.9        ] 

«    y 

^    y    z 

§_!=_6 

5_6_7^_104 

y    2 

X     y      z 

?-l  =  5 

?  +  L0_8  =  14.9 

Z        X 

y      z      X 

21.    l  +  l_l=a 

"1   * 

25. 

2+l_?=ol 

a:      y      2 

a:     y      2 

1-1+1=6 

?-2_2=.0 

a;     3/     z 

z     y 

1,1      1 

1,14^ 

-  + =  c 

--f--_    =0 

y     z      X 

' 

a;     2      3 

22.    5z  +  q/  =  a    ^                     26. 

«^  +  ^y  +  <^2:  =  «  ] 

az-\-cx=^h    \\    . 

ax  —  hy  —  cz=^h   \ 

ai/  -{-  bx  =  c  . 

ax-^cy-\-bz  = 

c\ 

^    2a:-y__3y  +  2z_a;-y-z 
3  4  5 


28. 


x-y  _y 


x-\-z 


x  —  a 


a  +  ^4-  ^ 


*  Subtract  from  the  sum  of  the  three  equations  each  equation  separately, 
f  Multiply  the  equation.s  by  a,  6,  and  c,  respectively,  and  from  the  sum 
of  the  results  subtract  the  double  of  each  equation  separately. 


CHAPTER  XII. 

Problems  producing  Simultaneous  Equations. 

197.  It  is  often  necessary  in  the  solution  of  problems 
to  employ  two  or  more  letters  to  represent  tlie  quantities  to 
be  found.  In  all  cases  the  conditions  must  be  sufficient 
to  give  just  as  many  equations  as  there  are  unknown  quan- 
tities employed. 

If  there  be  more  equations  than  unknown  quantities, 
some  of  them  are  superfluous  or  contradictory  ;  if  there  be 
less  equations  than  unknown  quantities,  the  problem  is  in- 
determinate or  impossible. 

(1)  When  the  greater  of  two  numbers  is  divided  by  the 

less  the  quotient  is  4  and  the  remainder  3 ;  and  when 
the  sum  of.  the  two  numbers  is  increased  by  38,  and 
the  result  divided  by  the  greater  of  the  two  numbers, 
the  quotient  is  2  and  the  remainder  2.  Find  the 
numbers. 

Let  X  =  the  greater  number, 

and  y  =  the  smaller  number. 

Then  ^^  =  4, 

y 

and        ^±X±38^^2. 

X 

From  the  solution  of  these  equations  x  =  47,  and  y  =  11. 

(2)  If  A  give  B  $10,  B  will  have  three  times  as  much  money 

as  A.  If  B  give  A  $10,  A  will  have  twice  as  much 
money  as  B.     How  much  has  each  ? 


PROBLEMS.  167 


Let  X  =  number  of  dollars  A  has, 

and  y  =  number  of  dollars  B  has. 

Then  y  +  10  =  number  of  dollars  B  has,  and  a;  —  10  =  number 
of  dollars  A  has  after  A  gives  1 10  to  B. 

.-.  y  +  10  =  3  (a;  -  10),  and  a;  +  10  =  2  (y  -  10). 

From  the  solution  of  these  equations,  a;  =  22  and  y  =  2Q. 

Therefore,  A  has  $22  and  B  $26. 

Exercise  LXXV. 

1.  The  sum  of  two  numbers  divided  by  2  gives  as  a  quo- 

tient 24,  and  the  difference  between  them  divided  by 
2  gives  as  a  quotient  17.     What  are  the  numbers  ? 

2.  The  number  144  is  divided  into  three  numbers.  When 

the  first  is  divided  by  the  second,  the  quotient  is  3 
and  the  remainder  2  ;  and  when  the  third  is  divided 
by  the  sum  of  the  other  two  numbers,  the  quotient  is 
2  and  the  remainder  6.     Find  the  numbers. 

3.  Three  times  the  greater  of  two  numbers  exceeds  twice 

the  less  by  10 ;  and  twice  the  greater  together  with 
three  times  the  less  is  24.     Find  the  numbers. 

4.  If  the  smaller  of  two  numbers  be  divided  by  the  greater, 

the  quotient  is  .21  and  the  remainder  .0057 ;  but  if 
the  greater  be  divided  by  the  smaller,  the  quotient 
is  4  and  the  remainder  .742.  What  are  the  num- 
bers? 

5.  Seven  years  ago  the  age  of  a  father  was  four  times  that 

of  his  son ;  seven  years  hence  the  age  of  the  father 
will  be  double  that  of  the  son.    What  are  their  ages? 

6.  The  sum  of  the  ages  of  a  father  and  son  is  half  what  it 

will  be  in  25  years;  the  difference  between  their 
ages  is  one-third  of  what  the  sum  will  be  in  20  years. 
What  are  their  ages  ? 


168  ALGEBRA. 

7.  If  B  give  A  $  25  they  will  have  equal  sums  of  money  ; 

but  if  A  give  B  §22,  B's  money  will  be  double  that 
of  A's.     How  much  has  each  ? 

8.  A  farmer  sold  to  one  person  30  bushels  of  wheat  and 

40  bushels  of  barley  for  $67.50;  to  another  person 
he  sold  50  bushels  of  wheat  and  30  bushels  of  barley 
for  $85.  What  was  the  price  of  the  wheat  and  of 
the  barley  per  bushel  ? 

9.  If  A  give  B  $5  he  will  then  have  $6  less  than  B  ;  but 

if  he  receive  $5  from  B,  three  times  his  money  will 
be  $20  more  than  four  times  B's.  How  much  has 
each? 

10.  The  cost  of  12  horses  and  14  cows  is  $1900;  the  cost 
of  5  horses  and  3  cows  is  $650.  What  is  the  cost  of 
a  horse  and  a  cow  respectively  ? 

Note  I.   A  fraction  of  which  the  terms  are  unknown  may  be  rep- 
resented by  -• 

Ex.  A  certain  fraction  becomes  equal  to'  -J  if  3  be  added  to 
its  numerator,  and  equal  to  -f-  if  3  be  added  to  its  de- 
nominator.    Determine  the  fraction. 
Let  I  =  the  required  fraction. 


By  the  conditions  ^ 


and 


2/  +  3       ' 

From  the  solution  of  these  equations  it  is  found  that 
x  =  G, 
2/  =  18. 
Therefore  the  fraction  =  y%. 

11.  A  certain  fraction  becomes  equal  to  2  when  7  is  added 
to  its  numerator,  and  equal  to  1  when  1  is  subtracted 
from  its  denominator.     Determine  the  fraction. 


PROBLEMS.  169 


12.  A  certain  fraction  becomes  equal  to  ^  when  7  is  added 

to  its  denominator,  and  equal  to  2  when  13  is  added 
to  its  numerator.     Determine  the  fraction. 

13.  A  certain  fraction  becomes  equal  to  -J-  when  the  denom- 

inator is  increased  by  4,  and  equal  to  J^  when  the 
numerator  is  diminished  by  15.  Determine  the  frac- 
tion. 

14.  A  certain  fraction  becomes  equal  to  f  if  7  be  added  to 

the  numerator,  and  equal  to  |  if  7  be  subtracted  from 
the  denominator.     Determine  the  fraction. 

15.  Find  two  fractions  with  numerators  2  and  5  respectively, 

whose  sum  is  1-^,  and  if  their  denominators  are  inter- 
changed their  sum  is  2. 

16.  A  fraction  which  is  equal  to  -|  is  increased  to  -^j  when 

a  certain  number  is  added  to  both  its  numerator  and 
denominator,  and  is  diminished  to  f  when  one  more 
than  the  same  number  is  subtracted  from  each.     De- 
termine the  fraction. 
Note  II.   A  number  consisting  of  two  digits  which  are  unknown 
may  be  represented  by  10  a;  +  y,  in  which  x  and  y  represent  the  digits 
of  the  number.     Likewise,  a  number  consisting  of  thi'ee  digits  which 
are  unknown  may  be  represented  by  100  x  +  lOy  +  2,  in  which  x,  y, 
and  2  represent  the  digits  of  the  number. 

For  example,  consider  any  number  expressed  by  three  digits,  as 
364.  The  expression  364  means  300  +  60  +  4 ;  or,  100  times  3  +  10 
tinie&  6  +  4. 

Ex.  The  sum  of  the  two  digits  of  a  number  is  8,  and  if  36 
be  added  to  the  number  the  digits  will  be  inter- 
changed.    What  is  the  number  ? 

Let  X  =  the  digit  in  the  tens'  place, 

and  y  =  the  digit  in  the  units'  place. 

Then  lOo;  +  y  =  the  number. 

By  the  conditions,     a;  +  y  =  8,  (1) 

and  lOx  +  y +  36  =  10y  +  x.  (2) 


170  ALGEBRA. 


From  (2),  9a; -9?/ =  -36. 

Divide  by  9,  x  —  y  =  —  4:. 

Add  (1)  and  (3),    .     2  a;  =  4, 

.■.x  =  2. 
Subtract  (3)  from  (1),  2  y  -  12, 

.•.y  =  6. 

Hence,  tlie  number  is  26. 

17.  The  sum  of  the  two  digits  of  a  number  is  10,  and  if  54 

be  added  to  the  number  the  digits  will  be  inter- 
changed.    What  is  the  number  ? 

18.  The  sum  of  the  two  digits  of  a  number  is  6,  and  if  the 

number  be  divided  by  the  sum  of  the  digits  the  quo- 
tient is  4.    What  is  the  number  ? 

19.  A  certain  number  is  expressed  hj  two  digits,  of  which 

the  first  is  the  greater.  If  the  number  be  divided  by 
the  sum  of  its  digits  the  quotient  is  7  ;  if  the  digits 
be  interchanged,  and  the  resulting  number  diminished 
by  12  be  divided  by  the  difference  between  the  two 
digits,  the  quotient  is  9.     What  is  the  number  ? 

20.  If  a  certain  number  be  divided  by  the  sum  of  its  two 

digits  the  quotient  is  6  and  the  remainder  3 ;  if  the 
digits  be  interchanged,  and  the  resulting  number  be 
divided  by  the  sum  of  the  digits,  the  quotient  is  4 
and  the  remainder  9.    What  is  the  number  ? 

21.  If  a  certain  number  be  divided  by  the  sum  of  its  two 

digits  diminished  by  2,  the  quotient  is  5  and  the  re- 
mainder 1  ;  if  the  digits  be  interchanged,  and  the 
resulting  number  be  divided  by  the  sum  of  the  digits 
increased  by  2,  the  quotient  is  5  and  the  remainder 
8.    Find  the  number. 

22.  The  first  of  the  two  digits  of  a  number  is,  when  doubled, 

3  more  than  the  second,  and  the  number  itself  is  less 
by  6  than  five  times  the  sum  of  the  digits.  What 
is  the  number  ? 


PROBLEMS.  171 


23.  A  number  is  expressed  by  three  digits,  of  which  the  first 

and  last  are  alike.  By  interchanging  the  digits  in 
the  units'  and  tens'  places  the  number  is  increased  by 
64 ;  but  if  the  digits  in  the  tens'  and  hundreds'  places 
are  interchanged,  9  must  be  added  to  four  times  the 
resulting  number  to  make  it  equal  to  the  original 
number.    What  is  the  number  ? 

24.  A  number  is  expressed  by  three  digits.    The  sum  of  the 

digits  is  21 ;  the  sum  of  the  first  and  second  exceeds 
the  third  by  3 ;  and  if  198  be  added  to  the  number, 
the  digits  in  the  units'  and  hundreds'  places  will  be 
interchanged.    Find  the  number. 

25.  A  number  is  expressed  by  three  digits.     The  sum  of 

the  digits  is  9 ;  the  number  is  equal  to  forty-two 
times  the  sum  of  the  first  and  second  digits ;  and  the 
third  digit  is  twice  the  sum  of  the  other  two.  Find 
the  number. 

26.  A  certain  number,  expressed  by  three  digits,  is  equal  to 

forty-eight  times  the  sum  of  its  digits.  If  198  be 
subtracted  from  the  number,  the  digits  in  the  units' 
and  hundreds'  places  will  be  interchanged ;  and  the 
sum  of  the  extreme  digits  is  equal  to  twice  the  mid- 
dle digit.     Find  the  number. 

Note  III.  If  a  boat  move  at  the  rate  of  x  miles  an  hour  in  still 
water,  and  if  it  be  on  a  stream  that  runs  at  the  rate  of  y  miles  an 
hour,  then 

X  -^y  represents  its  rate  down  the  stream, 
x  —  y  represents  its  rate  up  the  stream. 

27.  A  waterman  rows  30  miles  and  back  in  12  hours.      He 

finds  that  he  can  row  5  miles  with  the  stream  in  the 
same  time  as  3  against  it.  Find  the  time  he  was 
rowing  up  and  down  respectively. 


172  ALGEBRA. 


28.  A  crew,  which  can  pull  at  the  rate  of  12  miles  an  hour 

down  the  stream,  finds  that  it  takes  twice  as  long  to 
come  up  the  river  as  to  go  down.  At  what  rate  does 
the  stream  flow  ? 

29.  A  man  sculls  down  a  stream,  which  runs  at  the  rate  of 

4  miles  an  hour,  for  a  certain  distance  in  1  hour  and 
40  minutes.  In  returning  it  takes  him  4  hours  and 
15  minutes  to  arrive  at  a  point  3  miles  short  of  his 
starting-place.  Find  the  distance  he  pulled  down 
the  stream  and  the  rate  of  his  pulling. 

30.  A  person  rows  down  a  stream  a  distance  of  20  miles 

and  back  again  in  10  hours.  He  finds  he  can  row 
2  miles  against  the  stream  in  the  same  time  he  can 
row  3  miles  with  it.  Find  the  time  of  his  rowing 
down  and  of  his  rowing  up  the  stream ;  and  also  the 
rate  of  the  stream. 

Note  IV.  When  commodities  are  mixed,  it  is  to  be  observed  that 
the  quantity  of  the  mixture  =  the  quantity  of  the  ingredients ;  the 
cost  of  the  mixture  =  the  cost  of  the  ingredients. 

Ex.  A  wine-merchant  has  two  kinds  of  wine  which  cost 
72  cents  and  40  cents  a  quart  respectively.  How 
much  of  each  must  he  take  to  make  a  mixture  of  50 
quarts  worth  60  cents  a  quart  ? 

Let  X  =  required  number  of  quarts  worth  72  cents  a 

quart, 
and  y  =  required  number  of  quarts  worth  40  cents  a 

quart. 
Then,         72  a;  =  cost  in  cents  of  the  first  kind, 

4O3/  =  cost  in  cents  of  the  second  kind  of  wine, 
and  3000  =  cost  in  cents  of  the  mixture. 

.-.  X  +  y  =  50, 
72a;  +  40y  =  3000. 
From  which  equations  the  values  of  x  and  y  may  be  found. 


PROBLEMS.  173 


31.  A  grocer  mixed  tea  that   cost  him  42  cents  a  pound 

with  tea  that  cost  him  54  cents  a  pound.  He  had  30 
pounds  of  the  mixture,  and  by  selling  it  at  the  rate 
of  60  cents  a  pound,  he  gained  as  much  as  10  pounds 
of  the  cheaper  tea  cost  him.  How  many  pounds  of 
each  did  he  put  into  the  mixture  ? 

32.  A  grocer  mixes  tea  that   cost  him  90  cents  a  pound 

with  tea  that  cost  him  28  cents  a  pound.  The  cost 
of  the  mixture  is  $61.20.  He  sells  the  mixture  at 
50  cents  a  pound,  and  gains  $3.80.  How  many 
pounds  of  each  did  he  put  into  the  mixture  ? 

33.  A  farmer  has  28  bushels  of  barley  worth  84  cents  a 

bushel.  With  his  barley  he  wishes  to  mix  rye  worth 
$1.08  a  bushel,  and  wheat  worth  $1.44  a  bushel,  so 
that  the  mixture  may  be  100  bushels,  and  be  worth 
$1.20  a  bushel.  How  many  bushels  of  rye  and  of 
wheat  must  he  take  ? 

Note  V.  It  is  to  be  remembered  that  if  a  person  can  do  a  piece 
of  work  in  x  days,  the,  part  of  the  work  he  can  do  in  one  day  will  be 
represented  by  ^. 

Ex.  A  and  B  together  can  do  a  piece  of  work  in  48  days ; 
A  and  C  together  can  do  it  in  30  days ;  B  and  C  to- 
gether can  do  it  in  26-|  days.  How  long  will  it  take 
each  to  do  the  work  ? 

Let  X  =  the  number  of  days  it  will  take  A  alone  to  do  the  work, 

y  =  the  number  of  days  it  will  take  B  alone  to  do  the  work, 

and      z  =  the  number  of  days  it  will  take  C  alone  to  do  the  work. 

Then,  -,  -,  -,  respectively,  will  denote  the  part  each  can  do 

X    V    z  •  1 

^  in  a  day, 

and  -  +  -  will  denote  the  part  A  and  B  together  can  do  in  a  day, 
X     y 


but  —  will  denote  the  part  A  and  B  together  can  do  in  a  day. 


174  ALGEBRA. 


Therefore, 

i  +  l  =  A 

X     y     48 

(1) 

Likewise, 

l  +  l  =  i 

x^  z      30 

(2) 

and 

1113 

y'^2"26f~80 

(3) 

Add  (1),  (2),  and  (3), 

2     2     2          11 

x^y-^z          120 

(4) 

Multiply  (1)  by  2, 

2      2                 1 
~x^~y                24 

(5) 

Subtract  (5)  from  (4), 

2       1 
2~20 
.-.  z  =  40. 

Subtract  the  double  of 

.-.  y  =  80. 

Subtract  the  double  of 

(3)  from  (4),  ?  =  ^ 
.-.  a;  =  120. 

34.  A  and  B  together  earn  $40  in  6  days;  A  and  C  to- 

gether earn  $  54  in  9  days ;  B  and  C  together  earn 
$80  in  15  days.     What  does  each  earn  a  day? 

35.  A  cistern  has  three  pipes,  A,  B,  and  0.     A  and  B  will 

fill  it  in  1  hour  and  10  minutes ;  A  and  C  in  1  hour 
and  24  minutes ;  B  and  C  in  2  hours  and  20  minutes. 
How  long  will  it  take  each  to  fill  it  ? 

36.  A  warehouse  will  hold  24  boxes  and  20  bales ;  6  boxes 

and  14  bales  will  fill  half  of  it.  How  many  of  each 
alone  will  it  hold  ? 

37.  Two  workmen  together  complete  some  work  in  20  days ; 

but  if  the  first  had  worked  twice  as  fast,  and  the  sec- 
ond half  as  fast,  they  would  have  finished  it  in  15  days. 
How  long  would  it  take  each  alone  to  do  the  work  ? 

38.  A  purse  holds  19  crowns  and  6  guineas ;  4  crowns  and 

5  guineas  fill  ^  of  it.  How  many  of  each  alone  will 
it  hold? 


PROBLEMS.  175 


39.  A  piece  of  work  can  be  completed  by  A,  B,  and  0  to- 

gether in  10  days;  by  A  and  B  together  in  12  days; 
by  B  and  0,  if  B  work  15  days  and  C  30  days.  How 
long  will  it  take  each  alone  to  do  the  work  ? 

40.  A  cistern  has  three  pipes,  A,  B,  and  0.     A  and  B  will 

fill  it  in  a  minutes ;  A  and  C  in  5  minutes ;  B  and 
C  in  c  minutes.  How  long  will  it  take  each  alone  to 
fill  it? 

Note  VI.  In  considering  the  rate  of  increase  or  decrease  in  quan- 
tities, it  is  usual  to  take  100  as  a  common  standard  of  reference,  so 
that  the  increase  or  decrease  is  calculated  for  every  100,  and  there- 
fore called  per  cent. 

It  is  to  be  observed  that  the  representative  of  the  number  result- 
ing after  an  increase  has  taken  place  is  100  +  increase  per  cent ;  and 
after  a  decrease,  100  —  decrease  per  cent. 

Interest  depends  upon  the  time  for  which  the  money  is  lent,  as 
well  as  upon  the  rate  per  cent  charged ;  the  rate  per  cent  charged 
being  the  rate  per  cent  on  the  principal  for  one  year.     Hence, 

o-      1    •   ,        .      Principal  X  Rate  X  Time 

Simple  interest  = ^ , 

^  100 

where  Time  means  number  of  years  or  fraction  of  a  year. 

Amount  =  Principal  +  Interest. 

In  questions  relating  to  stocks,  100  is  taken  as  the  representative 
of  the  stock,  the  price  represents  its  market  value,  and  the  per  cent 
represents  the  interest  which  the  stock  bears.  Thus,  if  six  per  cent 
stocks  are  quoted  at  108,  the  meaning  is,  that  the  price  of  $100  of 
the  stock  is  $108,  and  that  the  interest  derived  from  $100  of  the 
stock  will  be  yf^  of  $100,  that  is,  $6  a  year.  The  rate  of  interest  on 
the  money  invested  will  be  |^|  of  6  per  cent. 

41.  A  man  has  $  10,000  invested.    For  a  part  of  this  sum  he 

receives  5  per  cent  interest,  and  for  the  rest  4  per 
cent;  the  income  from  his  5  per  cent  investment  is 
$50  more  than  from  his  4  per  cent.  How  much  has 
he  in  each  investment  ? 


176  ALGEBRA. 


42.  A  sum  of  money,  at  simple  interest,  amounted  in  6 

years  to  $26,000,  and  in  10  years  to  $30,000.  Find 
the  sum  and  the  rate  of  interest. 

43.  A  sum  of  money,  at  simple  interest,  amounted  in  10 

months  to  $26,250,  and  in  18  months  to  $27,250. 
Find  the  sum  and  the  rate  of  interest. 

44.  A  sum  of  money,  at  simple  interest,  amounted  in  vi 

years  to  a  dollars,  and  in  n  years  to  h  dollars.  Find 
the  sum  and  the  rate  of  interest. 

45.  A  sum  of  money,  at  simple  interest,  amounted  in  a 

months  to  c  dollars,  and  in  b  months  to  d  dollars. 
Find  the  sum  and  the  rate  of  interest. 

46.  A  person  has  a  certain  capital  invested  at  a  certain  rate 

per  cent.  Another  person  has  $1000  more  capital, 
and  his  capital  invested  at  one  per  cent  better  than 
the  first,  and  receives  an  income  $80  greater.  A  third 
person  has  $1500  more  capital,  and  his  capital  in- 
vested at  two  per  cent  better  than  the  first,  and  re- 
ceives an  income  $  150  greater.  Find  the  capital  of 
each,  and  the  rate  at  which  it  is  invested. 

47.  A  person  has  $12,750  to  invest.    He  can  buy  three  per 

cent  bonds  at  81,  and  five  per  cents  at  120.  Find 
the  amount  of  money  he  must  invest  in  each  in  order 
to  have  the  same  income  from  each  investment. 

48.  A  and  B  each  invested  $  1500  in  bonds ;    A  in  three 

per  cents  and  B  in  four  per  cents.  The  bonds  were 
bought  at  such  prices  that  B  received  $5  interest 
more  than  A.  Both  classes  of  bonds  rose  ten  per 
cent,  and  they  sold  out,  A  receiving  $  50  more  than 
B.     "What  price  was  paid  for  each  class  of  bonds  ? 


PROBLEMS.  177 


49.  A   person  invests   §10,000  in  three   per   cent  bonds, 

$16,500  in  three  and  one-half  per  cents,  and  has  an 
income  from  both  investments  of  $1056.25.  If  his 
investments  had  been  §2750  more  in  the  three  per 
cents,  and  less  in  the  three  and  one-half  per  cents, 
his  income  would  have  been  62^  cents  greater. 
What  price  was  paid  for  each  class  of  bonds  ? 

50.  The  sum  of  $2500  was  divided  into  two  unequal  parts 

and  invested,  the  smaller  part  at  two  per  cent  more 
than  the  larger.  The  rate  of  interest  on  the  larger 
sum  was  afterwards  increased  by  1,  and  that  of  the 
smaller  sum  diminished  by  1 ;  and  thus  the  interest 
of  the  whole  was  increased  by  one-fourth  of  its  value. 
If  the  interest  of  the  larger  sum  had  been  so  in- 
creased, and  no  change  been  made  in  the  interest  of 
the  smaller  sum,  the  interest  of  the  whole  would  have 
been  increased  one-third  of  its  value.  Find  the  sums 
invested,  and  the  rate  per  cent  of  each. 

Note  VII.  If  x  represent  the  number  of  linear  units  in  the  length, 
and  y  in  the  width,  of  a  rectangle,  xy  will  represent  the  number  of 
its  units  of  surface ;  the  surface  unit  having  the  same  name  as  the 
linear  unit  of  its  sides. 

51.  If  the  sides  of  a  rectangular  field  were  each  increased 

by  2  yards,  the  area  would  be  increased  by  220 
square  yards;  if  the  length  were  increased  and  the 
breadth  were  diminished  each  by  5  yards,  the  area 
would  be  diminished  by  185  square  yards.  What  is 
its  area  ? 

52.  If  a  given  rectangular  floor  had  been  3  feet  longer  and 

2  feet  broader  it  would  have  contained  64  square  feet 
more ;  but  if  it  had  been  2  feet  longer  and  3  feet 
broader  it  would  have  contained  68  square  feet  more. 
Find  the  length  and  breadth  of  the  floor. 


178  ALGEBRA. 


53.  In  a  certain  rectangular  garden  there  is  a  strawberry- 

bed  whose  sides  are  one-third  of  the  lengths  of  the 
corresponding  sides  of  the  garden.  The  perimeter  of 
the  garden  exceeds  that  of  the  bed  by  200  yards; 
and  if  the  greater  side  of  the  garden  be  increased  by 
3,  and  the  other  by  5  yards,  the  garden  will  be  en- 
larged by  645  square  yards.  Find  the  length  and 
breadth  of  the  garden. 

Note  VIII.   Care  must  be  taken  'to  express  the  conditions  of  a 
problem  with  reference  to  the  same  principal  unit. 

Ex.  In  a  mile  race  A  gives  B  a  start  of  20  yards  and  beats 
him  by  30  seconds.  At  the  second  trial  A  gives  B  a 
start  of  32  seconds  and  beats  him  by  9y\  yards. 
Find  the  rate  per  hour  at  which  each  runs. 

Let  X  =  number  of  yards  A  runs  a  second, 
and     y  =  number  of  yards  B  runs  a  second. 
Since  there  are  1760  yards  in  a  mile, 

=  number  of  seconds  it  takes  A  to  run  a 

^  mile, 

and  ~  =  number  of  seconds  B  was  running  in  the 

^  ^  first  and  second  trials,  respectively. 

Hence.  11^-1^  =  30. 
and       W_™^32. 

y         ^ 

The  solution  of  these  equations  gives  x  =  5{f  and  y  =  5^j. 

5i§-  1 

That  is,  A  runs  — ^,  or  — ,  of  a  mile  in  one  second  : 
1760'        300 
and  in  one  hour,  or  3600  seconds,  runs  12  miles. 
Likewise,  B  runs  10j^2T  iiiiles  in  one  hour. 

54.  In  a  mile  race  A  gives  B  a  start  of  100  yards  and  beats 

him  by  15  seconds.  In  the  second  trial  A  gives  B  a 
start  of  45  seconds  and  is  beaten  by  22  yards.  Find 
the  rate  of  each  in  miles  per  hour. 


PROBLEMS.  179 


55.  In  a  mile  race  A  gives  B  a  start  of  44  yards  and  beats 

him  by  51  seconds.  In  the  second  trial  A  gives  B  a 
start  of  1  minute  and  15  seconds  and  is  beaten  by  88 
yards.     Find  the  rate  of  each  in  miles  per  hour. 

56.  The  time  which  an  express-train  takes  to  go  120  miles 

is  y^j  of  the  time  taken  by  an  accommodation-train. 
The  slower  train  loses  as  much  time  in  stopping  at 
difierent  stations  as  it  would  take  to  travel  20  miles 
without  stopping ;  the  express-train  loses  only  half 
as  much  time  by  stopping  as  the  accommodation- 
train,  and  travels  15  miles  an  hour  faster.  Find  the 
rate  of  each  train  in  miles  per  hour. 

57.  A  train  moves  from  P  towards  Q,  and  an  hour  later  a 

second  train  starts  from  Q  and  moves  towards  P 
at  a  rate  of  10  miles  an  hour  more  than  the  first 
train;  the  trains  meet  half-way  between  P  and  Q. 
If  the  train  from  P  had  started  an  hour  after  the 
train  from  Q  its  rate  must  have  been  increased  by  28 
miles  in  order  that  the  trains  should  meet  at  the 
half-way  point.     Find  the  distance  from  P  to  Q. 

58.  A  passenger-train,  after  travelling  an  hour,  meets  with 

an  accident  which  detains  it  one-half  an  hour ;  after 
which  it  proceeds  at  four-fifths  of  its  usual  rate,  and 
arrives  an  hour  and  a  quarter  late.  If  the  accident 
had  happened  30  miles  farther  on,  the  train  would 
have  been  only  an  hour  late.  Determine  the  usual 
rate  of  the  train. 

59.  A  passenger-train  after  travelling  an  hour  is  detained 

15  minutes ;  after  which  it  proceeds  at  three-fourths 
of  its  former  rate,  and  arrives  24  minutes  late.  If 
the  detention  had  taken  place  5  miles  farther  on,  the 
train  would  have  been  only  21  minutes  late.  Deter- 
mine the  usual  rate  of  the  train. 


180  ALGEBRA. 


60.  A  man  bought  10  oxen,  120  sheep,  and  46  lambs.    The 

cost  of  3  sheep  was  equal  to  that  of  5  lambs ;  an  ox, 
a  sheep,  and  a  lamb  together  cost  a  number  of  dol- 
lars less  by  57  than  the  whole  number  of  animals 
bought;  and  the  whole  sum  spent  was  $2341.50. 
Find  the  price  of  an  ox,  a  sheep,  and  a  lamb,  respec- 
tively. 

61.  A  farmer  sold  100  head  of  stock,  consisting  of  horses, 

oxen,  and  sheep,  so  that  the  whole  realized  §11.75  a 
head ;  while  a  horse,  an  ox,  and  a  sheep  were  sold 
for  $110,  $62.50,  and  $7.50,  respectively.  Had  he 
sold  one-fourth  of  the  number  of  oxen  that  he  did, 
and  25  more  sheep,  he  would  have  received  the  same 
sum.  Find  the  number  of  horses,  oxen,  and  sheep, 
respectively,  which  were  sold. 

62.  A,  B,  and  C  together  subscribed  $100.     If  A's  sub- 

scription had  been  one-tenth  less,  and  B's  one-tenth 
more,  C's  must  have  been  increased  by  $  2  to  make 
up  the  sum ;  but  if  A's  had  been  one-eighth  more, 
and  B's  one-eighth  less,  C's  subscription  would  have 
been  $  17.50.     What  did  each  subscribe  ? 

63.  A  gives  to  B  and  C  as  much  as  each  of  them  has ;  B 

gives  to  A  and  C  as  much  as  each  of  them  then  has ; 
and  0  gives  to  A  and  B  as  much  as  each  of  them 
then  has.  In  the  end  each  of  them  has  $6.  How 
much  had  each  at  first  ? 

64.  A  pays  to  B  and  0  as  much  as  each  of  them  has ;  B 

pays  to  A  and  0  one-half  as  much  as  each  of  them  then 
has ;  and  0  pays  to  A  and  B  one-third  of  what  each  of 
them  then  has.  In  the  end  A  finds  that  he  has 
$1.50,  B  $4.16|,  C  $.58|.  How  much  had  each  at 
first? 


CHAPTER  XIII. 

Involution  and  Evolution. 

198.  The  operation  of  raising  an  expression  to  any  re- 
quired power  is  called  Involution. 

Every  case  of  involution  is  merely  an  example  of  multi- 
plication, in  which  the  factors  are  equal.     Thus, 
(2a«)2  =  2a»x2a»  =  4a«. 

199.  Any  power  of  a  power  of  a  number  is  obtained  by 
taking  the  product  of  the  exponents  of  the  powers.  The 
proof  of  this  law  of  exponents,  in  its  general  form,  is : 

(a"*)'*  =  a**  X  a"*  X  a**  X to  n  factors. 


ton  terms, 


Hence,  if  the  exponent  of  the  required  power  be  a  com- 
posite number,  it  may  be  resolved  into  prime  factors,  the 
power  denoted  by  one  of  these  factors  may  be  found,  and 
the  result  raised  to  a  power  denoted  by  another,  and  so  on. 
Thus,  the  fourth  power  may  be  obtained  by  taking  the  sec- 
ond power  of  the  second  power ;  the  sixth  by  taking  the 
second  power  of  the  third  power ;  the  eighth  by  taking  the 
second  power  of  the  second  power  of  the  second  power. 

200.  From  the  Law  of  Signs  in  multiplication  it  is  evi- 
dent that, 

I.    All  even  powers  of  a  number  are  positive. 

II.  All  odd  powers  of  a  number  have  the  same  sign  as 
the  number  itself. 


182  ALGEBRA. 

Hence,  no  even  power  of  any  number  can  be  negative ; 
and  of  two  compound  expressions  whose  terms  are  identical 
but  have  opposite  signs,  the  even  powers  are  the  same. 
Thus, 

{h-af=\-(a-b)Y  =  {a-h)\ 

201.  A  method  has  been  given,  §  83,  of  finding,  without 
actual  multiplication,  the  powers  of  binomials  which  have 
the  form  (a  ±  ^). 

The  same  method  may  be  employed  when  the  terms  of  a 
binomial  have  coefficients  or  exponents. 

(1)  {a-hf=^a^~-2>a^b-\-2>ab''-b\ 

(2)  {px'-2^/f, 

=  {ba^f  -  3  {Px^)\2f)  +  ^{bx^) {2ff  -  (2f)\ 
=  125  :r«  -  IbOxY  +  mx'y^  -  8f. 

(3)  (a-by  =  a^-4:a%  +  e,a'b^-4:ab^  +  b*. 

(4)  (x'-hjy, 

==(x^y-^x'y(iy)  +  6(a^y(},yy-4:x'(iyf+(hJy, 
=  x'-2x^j  +  ixY  -  ^a^f  +  j\y\ 

202.  In  like  manner,  a  polynomial  of  three  or  more  terms 
may  be  raised  to  any  power  by  enclosing  its  terms  in  par- 
entheses, so  as  to  give  the  expression  the  form  of  a  binomial. 
Thus, 

(1)    (a+b  +  cy=la+(b  +  c)]', 

=  a^  +  Sa'{b  +  c)  +  3a(b  +  cy+  (b  +  cf , 
=  a^  +  Sa^b  +  Sa^c  -i-Sab^-}-  6abc 

+  3a6-2  +  b^-]-Sb^c  +  3b(^-i-(^. 


INVOLUTION   AND    EVOLUTION.  183 

(2)    (x^-^x^'  +  Sx  +  Ay, 

=  {x^-2s^y  +  2(a^-2x')(3x  +  A)  +  (Sx-h4:y,  ' 

=  af'-4:3^  +  10x*-4:a^-1x'  +  24:X+lQ. 

EXEECISE  LXXVI. 
"Write  the  second  members  of  the  following  equations : 

1.  (ay=  11.   (2a'bc^y=  21.    (-Sa'b'cy  = 

2.  (x'f=  12.    (-5a^/f=      22.    (-3^/  = 

3.  (x'i/y=  13.    {-lw?na?yy  =  2Z,    (-5a^bx^y  = 

6.  (x  +  2f=  16.  (2x-ay=  26.  (l-a-a2)2^ 

7.  (x-2y=  17.  (3a:  +  2a)*=  27.  (2-3a;+4ar^/= 

8.  (x  +  ^y=  18.  (2:r-2/)*=  28.  {l-2x+a^y  = 

9.  (l  +  2a;)^=:  19.  {x?y-2xfy=  29.  (l-aj+a;^/^ 
10.  (2m-l/=  20.  {ah-Zy=  30.  (l+a;-f ^r^)*^ 

Evolution. 

203.  The  operation  of  finding  any  required  root  of  an 
expression  is  called  Evolution. 

Every  case  of  evolution  is  merely  an  example  of  factor- 
ing, in  which  the  required  factors  are  all  equal.     Thus, 

the  square,  cube,  fourth roots  of  an  expression  are  found 

by  taking  one  of  the  two^  three,  Jour equal  factors  of  the 

expression. 


184  ALGEBRA. 

204.  The  symbol  which  denotes  that  a  square  root  is  to 
be  extracted  is  ^^J  \  and  for  other  roots  the  same  symbol  is 
used,  but  with  a  figure  written  above  to  indicate  the  root, 
thus,  -J/,  -^j  etc.,  signifies  the  third  root,  fourth  root,  etc. 

205.  Since  the  cube  of  a^  =  a^,  the  cube  root  of  a®  —  o?. 
Since  the  fourth  power  of  2o?  =  2^a^,  the  fourth  root  of 

Since  the  square  of  abc  =  a^^V,  the  square  root  of  a^^V 

o-        ,,  nob      c^b^  ,-,  ,    f-a^Z*^      a5 

Since  the  square  oi  —  =  -5-5,  the  square  root  01  -„  „  =  — . 

xy     xr]f  x^if      xy 

Hence, .the  root  of  a  simple  expression  is  found  by  divid- 
ing the  exponent  of  each  factor  by  the  index  of  the  root,  and 
talcing  the  product  of  the  resulting  factors. 

206.  It  is  evident  from  §  200  that 

I.   Any  ev&n  root  of  a  positive  number  will  have  the 
double  sign,  ±. 

II.    There  can  be  no  even  root  of  a  negative  number. 

III.  Any  odd  root  of  a  number  will  have  the  same  sign  as 
the  number. 

Thus,-yjJ-|^  =  ±|^;  V- 27mW  =  -  3  m«' ; 


i 


4[16W'^  ,  2^ 


But  V—  :i^  is  neither  +  x  nor  —  x,  for  (+  x^  =  +  :r^,  and 
{:-xJ  =  ^:^. 

■  The  indicated  even  root  of  a  negative  number  is  called 
an  impossible,  or  imaginary,  number. 


INVOLUTION    AND    EVOLUTION. 


185 


207.  If  the  root  of  a  number  expressed  in  figures  is  not 
readily  detected,  it  may  be  found  by  resolving  the  number 
into  its  prime  factors.  Thus,  to  find  the  square  root  of 
3,415,104 : 


2^ 

3415104 

2« 

426888 

3^ 

53361 

7 

5929 

7 

847 

1 

121 

11 

3,415,104  =  2«x  3^x72x112. 
V3,415,104 -2^x3  x7  xll  =  1848. 


o-      vj  Exercise  LXXVII. 

Bimpiiiy : 

1.  V^^  ^  V4^^  ^64,  -Va^Y',  -s/Wa^¥?,  -\/-32a". 

2.  ^-1728c«^^/,   \/3375FP,  ■v'3111696  A*. 

3.  V5336UWV^.   .«/IMZ^,     «E 

-^      '    Af       34322-''     \7292*' 

4.  V25  a^h^c^  +  -v^S^V  -  </81  a*^;V  -  -^32^^^^. 

5.  A/27^y  X  -^2437?  X  Vl6^. 

When  a  =  1,  6  =  3,  a:  =  2,  y  =  6,  find  the  values  of : 

6.  4 V2^—  -s/ahxy  +  b^a^b^xy. 

7.  2a^^x-\-b-s/\2hy-]-^ahx^bxy. 


8.  Va2  +  2ai  +  52x  W  +  3a2^,-f  3a52  +  i3. 

9.  A/^r«  -  3  ^.2a  +  3  ha?  -  o?  -  Vi'  +  t^'  -  2a5. 


186  ALGEBEA. 


Squaee  Eoots  of  Compound  Expressions. 

208.  Since  the  square  of  a  +  ^  is  a^-{-1ab-\-  W,  the  square 
root  of  a^  +  2  a5  +  5^  is  a  +  h. 

It  is  required  to  find  a  method  of  extracting  the  root 
a-^-h  when  c? -\-^ah-\-}P'  is  given  : 

Ex.  The  first  term,  a,  of  the  root  is  obviously  the  square  root  of  the 
first  term,  a^,  in  the  expression. 

CJ?  -\-^ab  A-WXaA-h  ^^  ^^  ^  ^®  subtracted  from  the 

given  expression,  the  remainder  is 
-         2  2a6  +  J^   Therefore  the  second  term, 

"■"  6,  of  the  root  is  obtained  when  the 

^ab  -\r  o  first  term  of  this  remainder  is  di- 

vided by  2  a,  that  is,  by  double  the 
part  of  the  root  already  found.  Also,  since  2ah-\-h^={2a-\-h)h,  the  di- 
visor is  completed  by  adding  to  the  trial-divisor  the  new  term  of  the  root. 


a 


(1)    Find  the  square  root  of  25  ^i;^  -  20  x^y  +  4  x^. 

2bx'-20a^y-\-^xy\bx~2x'y 
25:^2 

-20a?y-\-^xy 


10x-2a^y 


The  expression  is  arranged  according  to  the  ascending  powers  of  a;. 

The  square  root  of  the  first  term  is  5x,  and  5  a;  is  placed  at  the 
right  of  the  given  expression,  for  the  first  term  of  the  root. 

The  second  term  of  the  root,  —2a^y,  is  obtained  by  dividing 
—  20a;^y  by  10  x,  and  this  new  term  of  the  root  is  also  annexed  to  the 
divisor,  10  a;,  to  complete^ the  divisor. 

209.  The  same  method  will  apply  to  longer  expressions, 
if  care  be  taken  to  obtain  the  trial-divisor  at  each  stage  of 
the  process,  by  doubling  the  part  of  the  root  already  found, 
and  to  obtain  the  complete  divisor  by  annexing  the  new  term 
of  the  root  to  the  trial-divisor. 


INVOLUTION   AND   EVOLUTION.  187 

Ex.    Find  the  square  root  of 

16x^-24:a^+25x*~20a^+10a^-4:x-j-l  \4.a^-SaP+2x-l 
l6x^ 


8a^-Sa^ 


-24:x'^-{-25x* 
-24  0;^+  9x* 


8a^-Q>x^+2x 


16x'-20a^+10ar' 


Sa^-Qx'-lAx-l 


-  8ar^+  6x^-4:x+l 


The  expression  is  arranged  according  to  the  descending  powers 
of  X. 

It  will  be  noticed  that  each  successive  trial- divisor  may  be  obtained 
by  taking  the  preceding  complete  divisor  with  its  last  term  doubled. 

Exercise  LXXVIII. 
Extract  the  square  roots  of : 

1.  a*  +  4:a^  +  2a^-4:a+l. 

2.  x*-2x^i/  +  ^xY-2^y^  +  y*. 

3.  4ri''-12a^a;  +  5aV  +  6aV  +  aV.. 

4.  9a;«-12:r3/+16.^V-24a;y  +  4y«  +  16:ry. 

5.  4a«+16c«+16aV-32aV. 

6.  4^-^  +  9 -30a;- 20 a;3  +  37a:«. 

7.  lQx*-16abx'+16b''x'  +  4:a^P-8ah^  +  U\ 

8.  a^  +  25x'+l0x^-4:x^-20a^+lQ>-24:X. 

9.  x^  +  Sxy-Ax^7/-4:xi/-{-8x^/-103^/-{-y^. 

10.  4-12a-lla^  +  5a2-4a^  +  4a«+14a^ 

11.  da^  —  Qah  +  SOac  +  Qad-i-P-lObc  —  2bd 

+  25(^-i-l0cd-{~d\ 


188  ALGEBRA. 


12.  25a;«  -  3l2;y  -\-Ua^f-  30:c^y  +  /  -  8:^y^  +  10^,^. 

13.  m^  —  A.wJ  +  10  m^  —  20  m^  —  44  m^ 

+  35mH46m2-40m  +  25. 

14.  x^—2^y  —  -x^y^-\-xif-\-if. 

15.  :r^-4a:3y  +  6:r2y2_6:ry3  +  5y^-^  +  ^. 
17.   1  +  4  +  10     20     25     24     16^ 


X         0^  X^  X^ 

^2       5  ^^       a  ^a^-  ^^-   "^^"^      12      3  +  9' 


Square  Eoots  oiuiiiia^^IETicAL  Numbers. 

210.  In  the  general  method  of  extracting  the  square  root 
of  a  number  expressed  by  figures,  the  first  step  is  to  mark 
off  the  figures  in  periods. 

Since  1  =  P,  100  =  10^,  10,000  =100^,  and  so  on,  it  is  evident  that 
the  square  root  of  any  number  bet'^een  1  and  100  hes  between  1  and 
10  ;  the  square  root  of  any  nun^jpr  bel:ween!f^lOO  and  10,000  hes  be- 
tween 10  and  100.  In  other  wq^s,  the  square  root  of  any  number 
expressed  by  one  or  two  figures.  iS:  a  iLumber  of  one  figure  ;  the  square 
J^  root  of  any  number  expressed  ^Ytliredox  four  figures  is  a  number  of 
— two  figures  ;  and  so  on. 

Ifj^herefore,  a  dot  be  placed  over  the  units'  figure  of  a  square  num- 
ber, and  over  every  alternate  figure,  the  number  of  dots  will  be  equal 
to  the  number  of  figures  in  .its  square  root. 

Find  the  square  root  of  3249. 

3249  (57  In  this  case,  a  in  the  typical  form  a^  +  2a'b  +  h^ 

25  represents  5  tens,  that  is,  50,  and  h  represents  7. 

107)  749  The  25  subtracted  is  really  2500,  that  is,  a^,  and 

749  the  complete  divisor,  2a  +  &,  is  2  x  50  -h  7  =  107. 


INVOLUTION   AND   EVOLUTION.  189 

211.  The  same  method  will  apply  to  numbers  of  more 
than  two  periods  by  considering  a  in  the  typical  form  to 
represent  at  each  step  the  part  of  the  root  already  found. 

It  must  be  observed  that  a  represents  so  many  tens  with  respect  to 
the  next  figure  of  the  root. 

Ex.    Find  the  square  root  of  5,322,249. 

5322249(2307 

4 
43)132 

129 

4607)32249 
32249 

212.  If  the  square  root  of  a  number  have  decimal  places, 
the  number  itself  will  have  twice  as  many. 

Thus,  if  .21  be  the  square  root  of  some  number,  this  number  will 
be  (.21)2  =  .21  X  .21  =  .0441 ;  and  if  .111  be  the  root,  the  number  will 
be  (.111)*  =  .111  X  .111  =  .012321. 

Therefore,  the  number  of  decimal  places  in  every  square  decimal 
will  be  even,  and  the  number  of  decimal  places  in  th^e  root  will  be 
half  as  many  as  in  the  given  number  itself 

Hence,  if  the  given  square  number  contain  a  decimal,  and  a  dot  be 
placed  over  the  units'  figure,  and  then  over  every  ■aUemate  figure  on 
both  sides  of  it,  the  number  of  dots  to  the  left  of  the  decimal  point 
will  show  the  number  of  integral  places  in  the  root,  and  the  number 
of  dots  to  the  right  will  show  the  number  of  decimal  places. 

Ex.   Find  the  square  roots  of  41.2164  and  965.9664. 

41.2164  (  6.42  965.9664  (  31.08 

36  9_ 

124)521  61)65 

496  61 

1282)2564  6208)49664 

2564  49664 

It  is  seen  from  the  dotting  that  the  root  of  the  first  example  will 
have  one  integral  and  two  decimal  places,  and  that  the  root  of  the 
second  example  will  have  two  integral  and  two  decimal  places. 


190  ALGEBRA. 


213.  If  a  number  contain  an  odd  number  of  decimal 
places,  or  if  any  number  give  a  remainder  wben  as  many- 
figures  in  the  root  have  been  obtained  as  the  given  number 
has  periods,  then  its  exact  square  root  cannot  be  found. 
We  may,  however,  approximate  to  its  exact  root  as  near  as 
we  please  by  annexing  ciphers  and  continuing  the  operation. 

Ex.   Find  the  square  roots  of  3  and  357.357. 


3.(1.732 

27)200 
189 

357.3570(18.903 

1 

28)257 
224 

343)1100 
1029 

369)3335 
3321 

3462)7100 
6924 

37803) 147000 
113409 

Exercise  LXXIX. 
Extract  the  square  roots  of : 

1.  ^120,409 ;  4816.36 ;  1867.1041 ;  1435.6521 ;  64.128064. 

2.  16,803.9369 ;  4.54499761 ;  .24373969  ;  .5687573056. 

3.  .9;  6.21;  .43;  .00852;  17;  129 ;  347.259. 

4.  14,295.387;  2.5;  2000;  .3;  .03;  111. 
5.^  .00111;  .004;  .005;  2;  5;  3.25;  8.6. 

6       1.     16.     1.0  0..     169  .     289  .     400 

'^'     2'i    ^;    ¥i    "ST  5    T2T;    TTS")    T5    TT- 

Cube  Roots  of  Compound  Expressions. 

214.  Since  the  cube  of  a  +  5  is  a^  +  3  a^^)  +  3  ab^  +  h^,  the 
cube  root  of  a^  +  3  a^h  +  2>  a¥ -\- h^  \^  a -{-  h. 

It  is  required  to  find  a  method  for  extracting  the  cube 
root  a  +  h  when  a^  +  3  a^6  +  3  a¥  +  b^  is  given : 


INVOLUTION   AND    EVOLUTION.  191 

(1)    Find  the  cube  root  of  a^  +  3  a^h  +  3  aV  +  h^. 


3a2 


+  3a5  +  ^' 


Za^-\-Zah-\-W' 


3a2^>  +  3a52  +  53 


The  first  term  a  of  the  root  is  obviously  the  cube  root  of  the  first 
term  a^  of  the  given  expression. 

If  a^  be  subtracted,  the  remainder  is  3a^5  +  ZaW  +  h^ \  therefore, 
the  second  term  h  of  tbe  root  is  obtained  by  dividing  the  first  term 
of  this  remainder  by  three  times  the  square  of  a. 

Also,  since  3  a^b  +  3  aJ'  +  b^=={3a^  +  Zab  +  &')  b,  the  complete  divisor 
is  obtained  by  adding  3  a6  +  6'  to  the  trial-divisor  3  a*. 

(2)    Find  the  cube  root  of  8a^  +  3Qx^7/  +  54:Xi/^  +  27y». 
12ar^  8r» 


(6a:+3y)  32/=__18£^4-9^ 


Ux'+lSxy+df 


36  a:«y+54  2^^+27/ 
36a;^y+54a:.y^+27.^/» 


The  cube  root  of  the  first  term  is  2x,  and  this  is  therefore  the  first 
terra  of  the  root. 

The  second  term  of  the  root,  3y,  is  obtained  by  dividing  36a:*y  by 
3  (2  a;)'  =  12  a;',  which  corresponds  to  3  a*  in  the  typical  form,  and  is 
completed  by  annexing  to  1 2  a;*  the  expression  ^3(2a;)  +  3y|3y  =  18a;y 
+  93/*,  which  corresponds  to  3a6  +  6*,  in  the  typical  form. 

215.  The  same  method  may  be  applied  to  longer  expres- 
sions by  considering  a  in  the  typical  form  3  a^  -{-  3  a5  +  i^  to 
represent  at  each  stage  of  the  process  ike  part  of  the  root 
already  found. 

Thus,  if  the  part  of  the  root  already  found  be  a;  +  y,  then  3  a'  of 
the  typical  form  will  be  represented  by  3  (x  +  yf ;  and  if  the  third 
term  of  the  root  be  +  z,  the  3  a5  +  &'  will  be  represented  by  3  (a;  +  y)  z 
+  2*.  So  that  the  complete  divisor,  3  a'  +  3  a6  +  J',  will  be  repre- 
sented by  3  (a;  +  y)*  +  3  (a;  +  3/)  2  +  2*. 


192                                                 ALGEBRA. 

Find  the  cube  root  of  x^  - 
3  a;* 

[x^-rc-l 

3a;*-3x3 

+    ar» 

-3a;5        +3^_    a^ 

(3a;^-3x-l)(-l)  = 

+  3ar^ 

-3a^ 

+  3x  +  l 

-3a;4  +  6a;-3-3a;-l 

Sx^-Gar^ 

+  3a;  +  l 

-3a;4  +  6a^-3a;-l 

The  root  is  placed  above  the  given  expression  for  convenience  of 
arrangement. 

The  first  term  of  the  root,  x^,  is  obtained  by  taking  the  cube  root 
of  the  first  term  of  the  given  expression ;  and  the  first  trial-divisor, 
3  a;'*,  is  obtained  by  taking  three  times  the  square  of  this  term  of  the 
root. 

The  first  complete  divisor  is  found  by  annexing  to  the  trial-divisor 
(3  a;'^  —  X)  (—  x),  which  expression  corresponds  to  (3  a  +  &)  Z>  in  the 
typical  form. 

The 'part  of  the  root  already  found  (a)  is  now  represented  by  a;^  — ar; 
therefore  3a^  is  represented  by  3  (a;^  —  a)*^  =  3  a^  —  6  a,-^  +  3  a;'',  the  sec- 
ond trial-divisor  ;  and  (3  a  -h  6)  6  by  (3x^  —  3  a;  —  1)  (—  1) ;  therefore, 
in  the  second  complete  divisor,  3  a^  4-  (3  a  -f  &)  6  is  represented  by 

(3a'4-Ga='  -|-  3a-2)  -f  (-3a;2-  3a;-l)  X  (-  1)  =  3  x""  -Q>x^  -|-  3  a;  +  1. 

Exercise  LXXX. 
Find  the  cube  roots  of : 

1.  a;3-f6r^y+12V+8y'-       3.  :^ +  I2x'' +  ^^x +  (j^. 

2.  a^- 9^2  + 27a -27.  4.  x^~2>a3^-{-ba^x^-^a'x-a\ 

5.  x^-\-2>x^  +  ^x^-^1a^  +  Qx'-\-?>x-\-l. 

6.  1- 9a;  +  392:2 _99^_l_  156^4  _  144^5 _|_(34^6_ 

7.  a«-6a^  +  9a^  +  4a3-9a2_6a-l. 

8.  64a;«  -f  192^^  -f  144:r^  -  2>2oi^  -  36:^^  +  12a;  -  1. 

9.  l-Zx-\-Q>x'-l0a?-\-l2x''-\2a^-\-l0x''-Q>x'+2>x^-x\ 


INVOLUTION   AND   EVOLUTION.  193 

10.  a«  -\-<da'b-  135  o^b^  +  729  ah'  -  729  h\ 

11.  6«  - 12  bc^  +  60  b^d"  - 160  bh^  +  240  iV  - 192  b'c  +  64  Z/«. 

12.  3  a«  +  48  a'b  +  60  a^^^  -  80  a%^  -  90  a26*+ 108 ab^-  2W. 

Cube  Boots  of  Arithmetical  Numbers. 

216.  In  extracting  the  cube  root  of  a  number  expressed 
by  figures,  the  first  step  is  to  mark  it  off  into  periods. 

Since  1  =  P,  1000=  lO*.  1,000,000  =  lOO^,  and  so  on,  it  follows  that 
the  cube  root  of  any  number  between  1  and  1000,  that  is,  of  any  num- 
ber which  lias  one,  two,  or  three  figures,  is  a  number  of  one  figure  ; 
and  that  the  cube  root  of  any  number  between  1000  and  1,000,000, 
that  is,  of  any  number  which  ha.8  four,  five,  or  six  figures,  is  a  number 
of  two  figures  ;  and  so  on. 

Hence,  if  a  dot  be  placed  over  every  third  figure  of  a  cube  num- 
ber, beginning  with  the  units'  figure,  the  number  of  dots  will  be  equal 
to  the  number  of  figures  in  its  cube  root. 

217.  If  the  cube  root  of  a  number  contain  any  decimal 
figures,  the  number  itself  will  contain  thi'ee  times  as  many. 

Thus,  if  .3  be  the  cube  root  of  a  number,  the  number  is  .3  x  .3  X  .3 
=  .027. 

Hence,  if  the  given  cube  number  have  decimal  places,  and  a  dot 
be  placed  over  the  units'  figure  and  over  every  third  figure  on  both 
sides  of  it,  the  number  of  dots  to  the  left  of  the  decimal  point  will 
show  the  number  of  integral  figures  in  the  root ;  and  the  number  of 
dots  to  the  right  will  show  the  number  of  decimal  figures  in  the  root. 

If  the  given  number  be  not  a  perfect  cube,  ciphers  may  be  an- 
nexed, and  a  value  of  the  root  may  be  found  as  near  to  the  true  value 
as  we  please. 

218.  It  is  to  be  observed  that  if  a  denote  the  first  term 
of  the  root,  and  b  the  second  term,  the  first  complete  divisor 

'^  3a'  +  3ab-\-b\ 

and  the  second  trial-divisor  is  3  (a  +  ^)^i  that  is, 

^a^  +  ^ab  +  U\ 


194 


ALGEBRA. 


wliich  may  be  obtained  from  the  preceding  complete  divisor 
by  adding  to  it  its  second  term  and  twice  its  third  term, 

3a2+6a^)  +  352 

a  method  which  will  very  much  shorten  the  work  in  long 
arithmetical  examples. 

219.    Ex.   Find  the  cube  root  of  14,102.327296. 

14102.327296  1 24.16 


8 


a3  =  23  = 

3a2  =  3(20/  = 

3a&  =  3(20x4)  = 

^»2  =  42  = 


252  = 

3a'2  =  3(240)2  = 

3a'S'  =  3x  240x1  = 

5'2  =  12  = 

3a'5'  = 
25'2=3 
3a"2  = 

3a''5''  =  3x  2410x6  = 
5"2  =  62  = 


It  will  be  observed  that  in  this  example  a  represents  20  and  h 
represents  4  ;  a',  10  (a+6),  represents  240  and  V  represents  1 ;  a",  10 
(a'+  5'),  represents  2410  and  5"  represents  6. 

It  will  be  observed,  also,  that  the  trial-divisors  are  formed  from 
the  preceding  complete  divisors,  according  to  the  method  explained 
in  §  218. 


1200 

6102 

240 

16 

1456 

5824 

240 

278327 

32 

172800 

720 

1 

173521 

173521 

720 

104806296 

2 

17424300 

43380 

36 

17467716 

104806296 

INVOLUTION   AND   EVOLUTION. 


195 


Exercise  LXXXI. 
Find  the  cube  roots  of : 


13.  33,076161. 

14.  102,503.232. 

15.  820.025856. 

16.  8653.002877. 

17.  1.371330631. 

18.  20,910.-518875. 


1.  274,625.  7.    1001.613. 

2.  110,592.  8.    1,259,712. 

3.  262,144.  9.   2.803221. 

4.  884.730.  10.   7,077,888. 

5.  109,215,352.       11.    12.812904. 

6.  1,481,544.  12.   56.623104. 
19.   91.398648466125.  20.   5.340104393239. 

21.  Find  to  four  figures  the  cube  roots  of  2.5 ;  .2 ;  .01 ;  4  ;  .4. 

220.  Since  the  fourth  power  is  the  square  of  the  square, 
and  the  sixth  power  the  square  of  the  cube ;  the  fourth  root 
is  the  square  root  of  the  square  root,  and  the  sixth  root  is 
the  cube  root  of  the  square  root.  In  like  manner,  the 
eighth,  ninth,  twelfth roots  may  be  found. 

Exercise  LXXXII. 
Find  the  fourth  roots  of: 

1.  81  a"  -  540  a^b  +  1350  a^V  -  1500  ah^  +  625  h\ 

2.  l_4a;+10:r2-16:r3+19^*-16a/'+10a;«-4a;^  +  a;«. 

Find  the  sixth  roots  of : 

3.  64-192a;  +  240a;2-160a;3+60a;^-12a--^-f  a;«. 

4.  729a;«- 1458.^ +  1215a;^-540:t'3+135a;2- 18a; +1. 

Find  the  eighth  root  of : 

5.  1  -  8y  +  2^if  -  mif  +  70y* -  SO?/'  +  28/ -  8/  +  /. 


CHAPTER  XIV. 

Quadratic  Equations. 

221.  An  equation  which  contains  the  square  of  the  un- 
known quantity,  but  no  higher  power,  is  called  a  quadratic 
equation. 

222.  If  the  equation  contain  the  square  only,  it  is  called 
a  pure  quadratic ;  but  if  it  contain  the  first  power  also,  it  is 
called  an  affected  quadratic. 

PUEE    QUADEATIC   EQUATIONS. 

Solve  the  equation  5^  —  48  =  2a.^. 

5  ^2  ___  ^Q  __  2  ^^  It  -vvill  be  observed  that  there  are  two  roots  of 

3  ^  __  ^g        equal  value  but  of  opposite  signs  ;  and  there  are 

2  __  1  /;.         only  two,  for  if  the  square  root  of  the  equation, 

.     .       3?  =  16,  were  written  ±  ic  =  ±  4,  there  would  be 

-^  only  two  values  of  x ;    since  the  equation  —  x 

=  +  4  gives  re  =  —  4,  and  the  equation  —  a;  =  —  4  gives  a;  =  4. 

Hence,  to  solve  a  pure  quadratic, 

Collect  the  unTcnown  quantities  on  one  side,  and  the  "known 
quantities  on  the  other ;  divide  hy  the  co-efficient  of  the  un- 
Tcnown quantity ;  and  extract  the  square  root  of  each  side 
of  the  resulting  equation. 

Solve  the  equation  3a;^  —  15  =  0. 

3ar  —  10  =  0  It  will  be  observed  that  the  square  root  of  5 

So.'^  ^=  15  cannot  be  found  exactly,  but  an  approximate 

s^  =:z^  value  of  it  to  any  assigned  degree  of  accuracy 

^'x=  ±V5  ^^^  ^®  found. 


QUADRATIC    EQUATIONS.  197 

223.  A  root  which  is  indicated,  but  which  can  be  found 
only  approximately,  is  called  a  Surd. 

Solve  the  equation  3a;^  +  15  =  0. 

oar  -f-lo  =  0  It  will  be  observed  that  the  square  root 

Zar  ==  —  15  of  —  5  cannot  be  found  even  approximately; 

^  =  —  5  for  the  square  of  any  number,  positive  or 

.  ^  -_  _|_-yd~5  negative,  is  positive. 

224.  A  root  which   is  indicated,  but  which  cannot  be 
found  exactly  or  approximately,  is  imaginary.     §  206. 

g^j^g.  Exercise  LXXXIII. 

1.  ^-3  =  46.  6.   6x''-9  =  2a^-{-2i. 

2.  2(:r2-l)-3(ar'+l)+14=0.  7.    (x  +  2y  =  ^x  +  5. 

3    ^-5  ,  2^+1,^1  a^      ^-1Q_7     50+a:^ 

•       3     "^       6  2'  5  15  25    * 

l-\-x'^l~x       *  •      x^'  +  S   "*"    x^-i-9 

^3         17  ,^    Q     ,  7      65a: 

''    4^-6^=3-  ''•   ^"  +  ^==— • 

11    4r^+5      2a:^-5_7ar^-25      ' 
*       10  15  20     * 

10a;^+17      12:r^  +  2_5a;^-4 
18  lla;2_g  9      • 

21  Sor^-ll        3  • 

14.   x^-\~hx-}-a  =  bx  (l—bx). 
15.    mar^-{-n  —  q.  16.    x^  — ax-\-b  =  ax  (x—1). 


198    •  ALGEBRA. 


Affected  Quadratic  Equations. 

225.  Since  (ax  db  hy  —  c^a?  ±  2  ahx  +  W,  it  is  evident  that 
tlie  expression  c^o^  dr2a5;r  lacks  only  the  third  term,  l?,  of 
being  a  complete  square. 

It  will  be  seen  that  this  third  term  is  the  square  of  the 
quotient  obtained  from  dividing  the  second  term  by  twice  the 
square  root  of  the  first  term. 

226.  Every  affected  quadratic  may  be  made  to  assume 
the  form  of  aV  ±  2  abx  =  c. 

The  first  step  in  the  solution  of  such  an  equation  is  to 
complete  the  square ;  that  is,  to  add  to  each  side  the  square 
of  the  quotient  obtained  from  dividing  the  second  term,  by 
twice  the  square  root  of  the  first  term. 

The  second  step  is  to  extract  the  square  'root  of  each  side 
of  the  resulting  equation. 

The  third  and  last  step  is  to  reduce  the  resulting  simple 
equation. 

(1)    Solve  the  equation  IQa? -\-bx  —  ?)=^lo(^  —  x-{-^b. 

16ar^  +  5a;  -  3  =  7a;2  -  a;  +  45. 
Simplify,  Oar' +  6  a;  =  48. 

Complete  the  square,  9ar'  +  6a;  +  l  =  49. 
Extract  the  root,  3a;  +  l  =  ±7. 

Keduce,  3a;  =  —  l  +  7or  —  1  —  7, 

3  a;  =  6  or  -  8, 
.-.   a;  =  2  or  -  2f . 

Verify  by  substituting  2  for  x  in  the  equation 

IGar' +  5a;  -  3  =  7ar*  -  a;  +  45, 
16  (2f  +  5  (2)  -  3  =  7  (2f  -  (2)  +  45, 
64  +  10-3  =  28-2  +  45, 
71  =  71. 


QUADRATIC   EQUATIONS.  199 

Verify  by  substituting  —  2f  for  x  in  the  equation 

16(-|f  +  5(-|)-3  =  7(-f)»-(-|)+45; 

^°^-¥-3  =  H^  +  |  +  45, 

1024  -  120  -  27  =  448  +  24  +  405, 

877  =  877. 


(2)    Solve  the  equation  3  rc^  —  4  a:  =  32. 

Since  the  exact  root  of  3,  the  coefficient  of  a;*,  cannot  be  found,  it 
is  necessary  to  multiply  or  divide  each  term  of  the  equation  by  3  to 
make  the  coefficient  of  cc*  a  square  number. 

Multiply  by  3,  9  a;*  - 12  a;  =  96. 

Complete  the  square,    9ic*  —  12a;  +  4  =  100. 
Extract  the  root,  3  x  —  2  =  ±  10. 

Reduce,  3a;  =  2  +  lOor  2- 10  ; 

3a;  =  12  or -8. 
.-.  a;  =  4  or  -  2f . 


Or,  divide  by  3, 

a?- 

4a;_ 
'  3  ' 

_32 
3' 

Complete  the  square, 

:>?- 

4a; 
3 

^1= 

32     4     100 
3      9       9  ' 

Extract  the  root. 

X 

2. 
3" 

;.  a;  = 

2  ±10 
3     ' 
=  4or-2f. 

Verify  by  substituting 

4  for 

X  in 

the  original  equation. 

48- 

-16  = 
32  = 

=  32, 
=  32. 

Verify  by  substituting  —  2f  for  x  in  the  original  equation, 
21J-(-10f)  =  32, 
32  =  32. 


200  ALGEBRA. 

(3)    Solve  the  equation  —  Sar^-{-bx  =  —  2. 

Since  the  even  root  of  a  negative  number  is  impossible,  it  is  necessary 
to  change  the  sign  of  each  term .     The  resulting  equation  is, 


\x^-5x  =  2. 

Multiply  by  3, 

9r^-15:r  =  6. 

Complete  the  square, 

9.^-15.+  25^49_ 
4       4 

Extract  the  root, 

s^~t4 

Reduce, 

-=¥• 

3aj  =  6or-l. 

.-,  ar  =  2or-|, 

Or,  divide  by  3, 

„     5a;      2 

Complete  the  square, 

^      5a;      25_49 
'3       36     36' 

Extract  the  root, 

6        6 

.■.^-'t\ 

=  2or-^. 

If  the  equation  3aj^  —  5a;==2be  multiplied  by /our  times  the  coeffi- 
cient of  ar^,  fractions  will  be  avoided : 

36a;2-60x  =  24. 
Complete  the  square,  36  a:*  — 60  a;  +  25  =  49. 
Extract  the  root,  6  a;  — 5  =  ±7, 

6a;  =  5±7, 
6a;=12or-2. 
.'.  a;=  2  or  —  ^. 

It  will  be  observed  that  the  number  added  to  complete  the  square 
by  this  last  method  is  the  square  of  the  coefficient  of  x  in  the  original 
equation  3  a;^  —  5  a;  =  2. 


QUADRATIC   EQUATIONS.  201 

3  1 

(4)  Solve  the  equation =  2. 

b—x     Zx—b 

Simplify  (as  in  simple  equations), 

4a;2_23x  =  -30. 
Multiply  by  four  times  the  coefficient  of  a?,  and  add  to  each 
side  the  square  of  the  coefficient  of  x, 

64ar»  -  (  )  +  (23)'  =  529  -  480  =  49. 
Extract  the  root,  8  a;  -  23  =  ±  7. 

Reduce,  8a;  =  23±7; 

8a?  =  30  or  16. 
.-.  a;  =  3|  or  2. 
If  a  trinomial  be  a  perfect  square,  its  root  is  found  by  taking  the 
roots  of  ih.Q  first  and  third  terms  and  connecting  them  by  the  sign  of 
the  middle  term.  It  is  not  necessary,  therefore,  in  completing  the 
square,  to  write  the  middle  term,  but  its  place  may  be  indicated  as  in 
this  example. 

(5)  Solve  the  equation  12a?—'^0x  =  ~1. 

Since  72  =  2^  X  3',  if  the  equation  be  multiplied  by  2,  the  coeffi- 
cient of  x*  in  the  resulting  equation,  144  x*  —  60  a;  =  —  14,  will  be  a 
square  number,  and  the  term  required  to  complete  the  square  will  be 
(Iff  ==  (I)'  =  ^-  Hence,  if  the  original  equation  be  multiplied  by 
4x2,  the  coefficient  of  a^  in  the  result  will  be  a  square  number,  and 
fractions  will  be  avoided  in  the  work. 

Multiply  the  given  equation  by  8, 

576  a:* -240  a;  =  -56. 
Complete  the  square,  576  x*  —  (  )  +  25  =  —  31. 
Extract  the  root,  24a;  — 5  =  ±V— 31. 

Reduce,  24  a;  =  5  ±  V^TsI. 

.-.  a;  =  ^5(5±Vir3i). 

Note.  In  solving  the  following  equations,  care  must  be  taken  to 
select  the  method  best  adapted  to  the  example  under  consideration. 


Solve :                   ^ 

Exercise  LXXXIV. 

1.   x'  +  4.x=^\2. 

4.  3?-lx==^.          7.   a^-x  =  Q>. 

2.   x'-Qx^U. 

5.    ^x'-4:x=l.        8.    5a;2-3^  =  2. 

3.   x'-Vlx+^^k. 

6.    I2a^+x-l^0.    9.    2a^-21x=l4:. 

202  ALGEBRA. 


10.  ^-_2^  +  _L=:0.  13.  ^-hl^2:r-l 

312  ^  +  4^  +  6 

11.  ^-^  =  2(^  +  2).  14.  -^ ^±1_=:__I. 

2      3         ^    ^    ^  ^  +  1      2(a;  +  4)  18 

io    So;  ,    4  _13  -.K       2  3,2 

1<^.      ■  ~p  -; —  — — -  .  i.O. 


4       3a;       6  x-1      x-2     x~4: 

16.   5x(x-S)-2(ar'-6)  =  (x  +  S)(x  +  A). 

Sx  5^    Sx"  23 

*  2(x+l)      8      a;2-l      4(a;-l)* 

18.  (x~2)(x-4:)-2(x-l)(x-S)=:0. 

19.  i(a;-4)-|(a;-2)  =  i(2a;  +  3). 

20.  -(3a;2-a;-5)-J(a:2-l)  =  2(a;-2)2. 
5  3 

2x  ,     3a; -50  _12a;+70 

•  15~^3(10  +  a;)  190      ' 

22        ^     _  15  -  7a;  ^      14a;  -  9_^a;^- 3 

•   a;2_i      8(l-a;)'  *  8a;-3        a;+l' 

2a;-l  ■  l_^2a;-3  .        a;  +  5  ^x-6 

'     x-1       6       a;-2'  '  2a;+l      a;-2" 

2a;  +  3  7-a;    ._7-3a; 


28. 


2(2a;-l)      2(a;  +  l)      4-3a; 


29.  12^-ll^  +  10a;-78^^  1^ 

8ar^-7a;  +  6  '         2 

30.  _6 1^  =  _7 8_^ 

x  —  1      x-\-b      x-{-l      x  —  b 


QUADEATIC    EQUATIONS.  203 

227.    Literal  quadratic  equations  are  solved  as  follows : 
(1)    Solve  the  equation  ax^ -\-hx^=c. 

Multiply  the  equation  by  4  a  and  add  the  square  of  &, 

4aV  +  (  )  +  ^2  _  4  ac  +  h^. 
Extract  the  root,  2  ax  +  6  =  ±  \/4  ac  +  W. 

Reduce,  2  aa;  =  —  6  ±  V'4  ac  +  6*. 

—  h  ±  v'4  ac  4-  6* 


2a 

(2)  Solve  the  equation  ac?a:  —  ac3?  —  bcx  —  5(i. 

Transpose  hex  and  change  the  signs, 

aca?  +  6cx  —  adx  =  bd. 
Express  the  left  member  in  two  terms, 

ac3?  +  {he  —  ad)  x  =  bd. 
Multiply  by  4  ac, 

4  aVa^  +  4  ac  (6c  —  ac?)  re  =  4  a&cc?. 
Complete  the  square, 

4  a'^cV  +  (  )  +  (&c  -  acZ )'  =  6 V  +  2  ahcd  +  a*d*. 
Extract  the  root,     2  a/:x  +  {he  —  ad)  =  ±  {he  +  ad ). 
Reduce,  2aca;  =  — (6c-a<i)±(6c +  ac?) 

=  2  aci  or  —  2  he. 

d  h 

.'.  X  =  -  or . 

c  a 

pq 

(3)  Solve  the  equation  par  —  px -\-  qoc^ -{•  qx  =    _r    . 

Express  the  left  member  in  two  terms, 

{p  +  q)a^-{p-q)x=^^. 

Multiply  by  four  times  the  coefficient  of  x*, 
4:  {p  +  qf  a^  —  4.  {p^  —  <f)x  =  4pq. 
Complete  the  square, 

4{p  +  qf^-{  )  +  {p-qf=p'  +  2pq  +  ^. 
Extract  the  root,  2{p +  q)x  —  {p  —  q)  =  ±  {p  +  q). 

Reduce,  ^{p  +  q)x  =  {p  —  q)±(p  +  q), 

=  2p  or  —2q. 

■    X--.    -^     or         ^ 
p-^q  p  +  q 

Note.  The  left-hand  member  of  the  equation  when  simplified 
must  be  expressed  in  two  terms,  simple  or  compound,  one  term  con- 
taining :>?,  and  the  other  term  containing  x. 


204  ALGEBRA. 


o  1     .  Exercise  LXXXV. 

1.  c(^-\-2ax=--a^.  14.   x^ -\~  ax  =  a -}-  x. 

2.  oi?=^^ax^"lo?.  15.   x^-\-ax^hx-^ab. 

3.  a?=^-^~Zmx.  16.    £  +  ^  =  ^  +  1 

4  a      X      b      X 

4.  ^_5^_^__M  =  0.  17.    ^   •       ^  ^   • 


2  2-  a;      a;  +  6      «      a  +  6 

18.   ^  +  ^-^  =  0. 


{x  +  af      {x-af  3  '    4       3a 

6.  cx  =  ax'-\-hx'--^,  19.   ^±^=:a  +  ^^. 

a  +  6  a;-3  x-{-^ 

7.  ^  +  ^'=2a^.  20.   m^-1^^^^^'-^'). 


C'  C  TYin 


12      •     .2 


8.    (a^  +  1)  a;  =  aa;^  +  a.   .       21.    {ax  —  5)  {hx  —  a)  =  c^. 
x  —  a      x  —  b      x  —  c  bx-\-  a      nx  —  m 


1  1,1.1 


Tn.  ni 


10.  _^-_=i  +  ±  +  ±.     23.  -^i:;_  +  _::i_  =  c. 

a  +  o  +  ^      a      0      X  TYi-f-x      m  —  x 

^^    _1 l_=l±j^.    24.   (^-iy^'+2(3a-l)^_i^ 

'   a  —  x      a-\-x      a^  —  oc^'        '  4  a— 1 

^2^   ^?+2aM^^±^^2^       25.    (^^-^^)(^+l)^2r.. 

^2^52  ^•^-  ^-  ^2_|_^2 

13.      (2^-^y        5^  26.   4=i^  =  (m-^)l 

2  ,  a-5_14a^-5a^>-105^  ■  (2a~Sb)x 
'   "^  '^   ab^  I8a'b^  ^        2ab       ' 


QUADRATIC   EQUATIONS. 


205 


28.    ah3?- 


c 


29. 


m' 


c" 
4a2 


3m  — 2a      4<x  — 6m 

{a±hf 


30.    6:r  + 


b{a-b) 


2b  ah 


6.^ 


31. 
32. 
33. 
34. 
35. 

36. 

37. 

38. 


40. 

41. 
42. 
43. 

44. 
45. 


I  (V  4-  a^  +  ah)  =  ^x  (20a  +  4Z>). 

x^  —  (h  —  a)  c  =^  ax  —  hx  -\-  ex. 

01?  —  2mx  =  {n  —p  -|-  m)  (n  —p  —  m). 

.T^  —  (m  -f  w)  ^  =  i  (i?  +  g'  +  w  +  ^)  (p  +  <7  —  "A^  —  ■?^). 

7)1713?  —  (m.  +  '^)  (^^  +  1)  ^  +  (^i  +  ^)^  =  0. 

2h  —  x  -2a  ,  45  — 7a ^ a;  — 4a 
6a:  ax  —  hxah~  W 

20?  (a'-h')  -  (3a2  +  62)  (x-l)  =  (Sh'^  +  a')  (x+l). 

a  —  2h  —  x        bh  —  x     ,  2a  — a:  — 196 


46^ 


ax-{-2hx         2hx  —  ax 


=  0. 


a;+13a  +  36      j^_a-26 
ba-^h-x  x-\-2h' 


a:4-35 


36 


a  +  36 


8a2-12a6      962-4a2      (2a  +  36)(a;-36) 

713?  -\-px  —po?  —  \nx  -\-7n  — 71=^0. 

(a  +  6  +  c)r^-(2a  +  6  +  c)a:  +  a  =  0. 

(ax  —  h){c  —  d)  =  (a  —  6)  (ex  —  d)  x. 

2:fc-  +  l      1/1      2\      3a;+l 


0. 


f  +  1      lA      2\ 
6  x\b      aj 


1 I 1  ^       a       _  26a: +  6 

2ar^  +  a:-l      2ar^-3a;4-l      26a:-6       ax^-a' 


206  ALGEBRA. 


228.  An  affected  quadratic  may  be  reduced  to  the  form 
x^  -\-px  -{-  q  —  0,  in  which  p  and  q  represent  any  numbers, 
positive  or  negative,  integral  or  fractional. 


Ex.  Solve:  a? -{-px-\- q  =  0. 

4:x'-{)+p^=p^-^.q, 


2a;  — p  =  rb  Vp^  —  4  g-, 

.  By  this  formula,  the  values  of  x  in  an  equation  of  the 
form  o(^-\rpx-\~q  —  0,  may  be  written  at  once.  Thus,  take 
the  equation 

3a;2-5a;+2-=0.  • 

Divide  by  3,      or^  -  f  a;  +  f  -  0. 

Here,  P  =  ~"  f  >  ^^^  2'  =  f  • 

=  1  or  f . 

229.  A  quadratic  which  has  been  reduced  to  its  simplest 
form,  and  has  all  its  terms  written  on  one  side,  may  often 
have  that  side  resolved  hy  inspection  into  factors. 

In  this  case,  the  roots  are  seen  at  once  without  com- 
pleting the  square. 

(1)    ^o\YQa?-\-1x-m  =  0. 

Since  ar^  +  7a;  -  60  =  (a;  +  12)  (a;  -  5), 

the  equation  a^  +  7a;  —  60  =  0 

may  be  written    {x  +  12)  (a;  —  5)  =  0. 

It  will  be  observed  that  if  either  of  the  factors  a;  +  12  or  a;  —  5  is  0, 
\hQ  product  of  the  two  factors  is  0,  and  the  equation  is  satisfied. 

Hence,  a;  +  12  =  0  and  a;  —  5  =  0. 

:.x  =  —  12,  and  «  =  5. 


QUADRATIC   EQUATIONS.  207 

(2)    Solve  :i;2+ 7a;  =  0. 

The  equation  ar*  +  7x  =  0 

becomes  a;  (a;  +  7)  =  0, 

and  is  satisfied  if  x  =  0,  or  if  re  +  7  =  0. 

.•.  the  roots  are  0  and  —  7. 

It  will  be  observed  that  this  method  is  easily  applied  to  an  equation 
all  the  terms  of  which  contain  x. 


(3)    ^o\\Q2a?-3(^-e>x  =  0. 

The  equation  2a^  -a;'  —  6a;  =  0 

becomes  x  {2  a?  —  x  —  Q)  =  0, 

and  is  satisfied  if  a:  =  0,  or  if  2  a;*  —  a;  ~  6  =  0. 

By  solving  2a^— x— 6  =  0  the  two  roots  2  and  —  f  are  found. 
.'.  the  equation  has  three  roots,  0,  2,  —  f . 


(4)    Solve  rr3  +  ar2-4a:-4  =  0. 


The  equation 

x^  +  x^ 

-4ar- 

-4  = 

=  0 

becomes 

x^ix  +  l)- 

■4(a;  + 

1)  = 

=  0, 

=  (a:*.- 

4)(a:  + 

1)  = 

=  0. 

.-.  the  roots  of  the 

equation  are  —  1 

,2,-2 

(5)    Solve  k'-B.i-^- liar +12  =  0. 

Since  .^-2x'-ll.  +  12,^_^_j3^ 

a;  —  1 
the  equation  a^  —  2^2  —  lla;  +  12  =  0 

may  be  written  {x  —  1)  (o,^  —  .t  —  12)  =  0. 

The  three  roots  are  found  to  be  1,  —  3,  4. 

An  equation  which  cannot  be  resolved  into  factors  by  inspection 
may  sometimes  be  solved  by  guessing  at  a  root  and  reducing  by  divi- 
sion. In  this  case,  if  a  denote  the  root,  the  given  equation  (all  the 
terms  of  the  equation  being  written  on  one  side),  may  be  divided  by 
x~a. 


208  ALGEBRA. 


ExEECisE  LXXXVL 
Find  the  roots  of : 

1.  (x+l)(x-2')(x^+x~2')  =  0.  7.  x^-ar'-x  +  l  =  0. 

2.  (x''-Sx-{-2)(x--x-12)  =  0.  8.8:^-1  =  0. 

3.  (x-{-l)(x-2)(x+S)  =  -6.  9.  8.r'^+l  =  0. 

4.  2a^  +  4:x'-10x  =  0.  10.  a.''- 1  =  0. 

5.  (x^-x-6)(x'-x-20)  =  0.  11.  x(x-a)(x'-P)=^0. 

6.  x(x-]-l)(x-i-2)={a-i-2)(a+l)a.  12.  n(:^+l)+x-j-l=0. 

230.    If  r  and  /  represent  two  values  of  x,  then 

X  —  7^  =  0, 

and  a;  —  r'  =  0, 

.'.  (:r  —  r)  (a;  —  /)  =  0. 

This  is  a  quadratic  equation,  as  may  be  seen  by  performing  the 
indicated  multiplication. 

Now  r  and  r'  are  roots  of  this  equation ;  for,  if  either  r  or  r'  be 
written  for  x,  one  of  the  factors,  x  —  r,  x  —  r',  is  equal  to  0,  and  the 
equation  is  satisfied.  Also  r  and  r'  are  the  only  roots,  for  no  value 
of  X,  except  r  and  r',  can  make  either  of  these  factors  equal  to  0. 

Since  r  and  r'  may  represent  the  values  of  x  in  any  quadratic 
equation,  it  follows  that  every  quadratic  equation  has  two  roots,  and 
only  two. 

Again,  if  r,  r',  r",  represent  three  values  of  x, 
then,  {x  ~  r)  {x  -  r')  {x  -  r")  =  0. 

This  is  a  cubic  equation,  as  may  be  seen  by  performing  the  indi- 
cated multiplication.  Hence,  it  may  be  inferred  that  a  cubic  equa- 
tion has  three  roots,  and  only  three;  and  so,  for  any  equation,  that 
the  number  of  roots  is  equal  to  the  degree  of  the  equation. 

It  may  also  be  inferred  that  if  r  be  a  root  of  an  equation,  x  —  r 
ivill  be  a  factor  of  the  equation  when  the  equation  is  written  with  all 
its  terms  on  one  side. 


QUADRATIC   EQUATIONS.  209 

If  r  and  r'  represent  the  roots  of  the  general  quadratic  equa- 
tion, 3i^+px  +  q  =  0. 

This  equation  may  be  written  {x  —  r){x  —  r')  =  0, 
or,                                                 3?—  [r  +  r')  X  +  rr'  =  0. 

A  form  which  shows  that 
the  sum  of  the  roots  =  — p, 

and  the  product  of  the  roots  =  q. 

231.  It  will  be  seen  from  §  230  that  an  equation  may  be 
formed  if  its  roots  be  known. 

If  the  roots  of  an  equation  be  —  1  and  \, 
the  equation  will  be  (a;  +  1)  (a;  —  i)  =  0, 

4      4 
or,  by  multiplying  by  4,  4  a^  +  3  x  —  1  =  0. 

If  the  roots  of  an  equation  be  0,  1,  5, 
the  equation  will  be  (a;  —  0)  (a;  —  1)  (x  —  5)  =  0 ; 

that  is,  a;  (a;  —  1)  (a;  —  5)  =  0, 

or,  a:'  — 6ar*  +  5a;  =  0. 

If  X  occur  in  every  term  the  equation  will  be  satisfied  by  putting 
a;  =  0,  and  may  be  reduced  to  an  equation  of  the  next  lower  degree 
by  dividing  every  term  by  x. 

232.  By  considering  the  roots  of  a?  -\-px  +  5'  =  0," 
namely,  r  =  — -^  +     V/>^  —  4  5', 

and  /  =  — -^  — -  ^p^  —  4iq, 

it  will  be  seen  that  tbe  character  of  the  roots  of  an  equation 
may  be  determined  without  solving  it : 

I.  As  the  two  roots  have  the  same  expression,  Vp^  —  4  ^, 
both  roots  will  be  real,  or  both  will  be  imaginary. 

If  both  be  real,  both  will  be  rational  or  both  surds,  accord- 
ing as  p^  —  4  </  is  or  is  not  a  perfect  square. 


210  ALGEBRA. 

II.  When  p^  is  greater  than  4  q,  the  two  roots  will  be  real, 
for  then  the  expression  p^  —  ^q  is  positive,  and  therefore 
Vp  —  4  5-  can  be  found  exactly  or  approximately. 

Since  also  its  value  in  one  root  is  to  be  added  to  — -^, 

and  in  the  other  to  be  subtracted  from  — -,   the  two  roots 
will  be  different  in  value. 

III.  When  p^  is  equal  to  4  q,  the  roots  will  be  equal  in 
value. 

IV.  When  p^  is  less  than  4  q,  the  roots  will  be  imaginary, 
for  then  the  expressions^— 4  g  will  be  negative,  and  therefore 
Vp^  —  4:q  represents  the  even  root  of  a  negative  number,  and 
is  imaginary. 

V.  If  q  (=r  X  /)  be  positive,  the  roots,  if  real,  will  have 
the  same  sign,  but  opposite  to  that  of^  (since  r+/  =  — p). 

But  if  q  be  negative,  the  roots  will  have  opposite  signs. 


233.  Determine  by  inspection  the  character  of  the  roots  of: 

(1)  rr2-5a;  +  6  =  0. 

In  this  equation  p  is  ~  5,  and  q  is  6. 

.-.  \//-4^=V25-24  =  1. 
.•.  the  roots  will  be  rational,  and  hoih positive. 

(2)  x'+?>x+l  =  0. 

In  this  equation,  p  is  3,  and  g'  is  1. 

.-.  V/  -  4  g  =  V9  -  4  =  V5. 
.•.  the  roots  will  be  surds,  and  both  negative. 

(3)  a;2  +  3a7  +  4  =  0. 

In  this  equation  p  is  3,  and  g'  is  4, 

.-.  Vp2  _  4  ^  =  V9316  ==:\/i:7. 
.*.  the  roots  will  be  impossible. 


quadratic  equations.  211 

Exercise  LXXXVII. 
Form  the  equations  whose  roots  are : 

1.  2.  1.  5.   -5,-f  9.   0,~if,-l. 

2.  7,-3.  6.   --J,f  10.   a  —  2b,2>a-\-2b. 

3.  11  7.    3, -3,  f, -f.      11.    2a-b,h-Za. 

4.  I,  — f.  a   0,1,2,3.  12.    a(a+l),  1-a. 
Determine  by  inspection  the  character  of  the  roots  of : 

13.  x'-lx-\-l2  =  0.  17.  a^-{-4:X-\-l  =  0. 

14.  x'-1x-^0  =  0.  la  :r2-2:r  +  9  =  0. 

15.  a;2_|_4^_5=:0.  19.  3a,-2  — 4a:-4  =  0. 

16.  bx^-\-^  =  0.  20.  5r^  +  4a;  +  4  =  0. 

234.    It  is  often  useful  to  determine  the  maximum  or  mini- 
mum value  of  a  given  quadratic  expression. 

(1)    Find  the  maximum  or  minimum  value  of  1  +  ^  —  ^. 
Let  1  +  a;  —  x*  =  m ; 

then,  a^  —  a;  =  1  —  m, 

and  4a;'  — 0 +  1  =  5 -4m, 

2a;  — 1  =  ±\/5  — 4m. 


.'.  x  =  ^  ±  J  v5  — 4m. 
Now,  for  all  possible  values  of  a;,  5  —  4m  cannot  be  negative ;  that 
is,  m  cannot  be  greater  than  f ;  and  for  this  value  x  is  ^.     Therefore, 
^  is  the  maximum  value  of  the  given  expression. 

(2)    Find  the  maximum  or  minimum  value  of  o.-^  +  3:r  +  4. 
Let  a;'  +  3a;  +  4  =  m; 

then,  a;'  +  3a;  =  m  —  4, 

and,  4a;*  +  ( ) +  9  =  4m  — 7^__ 

2a;  +  3  =  ±V4m-7. 

a;  =  —  f  ±  J  V4m  — 7. 
For  all  possible  values  of  x,  4m  — 7  cannot  be  negative;  that  is, 
m  cannot  be  less  than  | ;  and  for  this  value  a;  =  —  f .     Therefore,  |  is 
the  minimum  value  of  the  given  expression. 


212  ALGEBRA. 


Exercise  LXXXVIII. 

Find  the  maximum  or  minimum  value  (and  determine 
which)  of: 

1.  4c  +  6x  —  x\         4.    (a~x)(x-h).         7.   ci^  —  2x-i-9. 

2.  i^±^.  5.    -^-.  8.    ^ . 

X  l-\-x^  (x-\-a)(x—b) 

3.  ^!±1.  6.   ^^  +  8^  +  20.  9.   -^. 

X  a-j-  XT 

10.  Divide  a  line  20  in.  long  into  two  parts  so  that  the  sum 

of  the  squares  on  these  two  parts  may  be  the  least 
possible. 

11.  Divide  a  line  20  in.  long  into  two  parts  so  that  the  rect- 

angle contained  by  the  parts  may  be  the  greatest 
possible. 

12.  Find  the  fraction  which  has  the  greatest  excess  over  its 

square. 

235.    Two  other  cases  of  the  solution  of  equations  b^  com- 
pleting the  square  should  be  noticed. 

I.  When  an]/  two  powers  of  x  are  involved,  one  of  which 
is  the  square  of  the  other. 

II.  When  the  addition  of  a  number  to  an  equation  of  the 
fourth  degree  will  make  both  sides  complete  squares. 

(1)    Solve8a;^  +  63.i;«=8. 

In  this  equation  the  exponent  6  is  the  double  of  3,  hence  x^  is 
the  square  of  a^. 

Sa^  +  eSx-^^S, 
256a«  +  ()  + (63)2  =  4225, 
16a^  +  63  =  ±65, 

16x3  =  2,  or -128, 
ar»  =  i,  or  -  8. 
By  taking  cube  root,  re  =  |,  or  —  2. 


QUADRATIC   EQUATIONS. 


213 


The  other  roots  of  the  equation  are  found  by  finding  the  remain- 
ing roots  of  the  equations,  x^  =  \,  and  a:^  =  —  8. 


Since,  P  =  i  .-.8x3-1  =  0 

Now,  by  \  230, 

8a;3  _  1  =  (2a;  -  1)  (4ar^  +  2x  +  1) 

.-.  (2a;-l)(4a^  +  2x  +  l)  =  0 

and  is  satisfied  if  4a^  +  2x4-1  =  0 

as  well  as  if  2  x  -  1  =  0 

The  solution  of  4x2  -h  2x  +  1  =  0 
gives      X  =  ^  (—  1  ±  V^). 


Since,  r'  =  -8,       .•.x3  +  8  =  0 

Now,  by  \  230, 

x3  +  8  =  (x  +  2)  (x»  -  2x  +  4) 

.•.(x  +  2)(x^-2x  +  4)  =  0 

and  is  satisfied  if  x*  —  2x  —  4  =  0 

as  well  as  if  x  -l-  2  =  0 

The  solution  of  a'-*  —  2x  -i-  4  =  0 
gives         x=l±\/^^. 


.*.  the  roots  are  J,  —  2,  1  iV—  3,  \{—\  ±  V—  3). 


(2)    Solve  a:*  -  10:r3  + 35^:2  _  50^  ^24  =  0. 
Take  the  square  root  of  the  left  side. 


2r^-5a; 


-10:^8  +  35^ 

-10a;«4-25:L'2 

10a;2_5o^_|_25 


2ar^-10a;4-5 


It  is  now  seen  that  if  1  were  added  the  square  would  be  complete, 
and  the  equation  would  be 

a;4  _  105.-3  -f  35 a;2  _  50^  -f  25  =  1. 
Extract  the  square  root  and  the  result  is, 

X*  —  5x  +  5  =  ±l. 
That  is,  x2  -  5  X  =  -  4,  or'-  6, 

4x2-0  +  25  =  9,  or  1, 

2x-5  =  ±  3,  or  ±  1, 
2x  =  8,  2,  6,  or  4, 
.-.  X  =  4,  1,  3,  or  2. 


214  ALGEBRA. 

Exercise  LXXXIX. 
Find  the  possible  roots  of: 

1.  x^+7x^  =  8.  8.  (x^~-9y=^3  +  ll(x'~2). 

2.  x^-5ar^-{-4:  =  0.  9.  ^•6-|-l4^_}_24  =  0. 

3.  S1x'-~9  =  4:x\  10.  19x^-}-2Wx'  =  x. 

4.  16a;«  =  172;*-l.  11.  :r«  + 22a;*  + 21  =  0. 

5.  32:r^°-33a;«+l  =  0.  12.  or^"*  +  3 ^r"' - 4  =  0. 

6.  (a:2_2)2=^(^_|_12).  13.  4.x'~20x^+23x'+5x  =  6. 

7.  a;''"-  — -— =0.  14.    -L  +  l-20-a 

3         12  ^"     ^" 

15.  :r*  -  4^:^  -  10:^2 +  28:r- 15  =  0. 

16.  x*-2a^-13x^i-Ux  =  -24:. 

17.  108:r''  =  20a;(9a.^-l)-51a;2+7. 

18.  (x'-l){:^-2)-\-(a^-S')(ix^-4:)  =  x*i-5. 

Problems  Involving  Quadratics. 

236.  Problems  whicli  involve  quadratic  equations  have 
apparently  two  solutions,  as  a  quadratic  has  two  roots. 
Sometimes  both  will  be  solutions ;  but  generally  one  only 
will  be  a  solution,  and  the  other  be  inconsistent  with  the 
conditions  of  the  problem.  No  difficulty  will  be  found 
in  selecting  the  result  which  belongs  to  the  problem,  and 
sometimes  a  change  may  be  made  in  the  statement  of  a 
problem  so  as  to  form  a  new  problem  corresponding  to  the 
solution  which  was  inapplicable  to  the  original  problem. 


QUADRATIC   EQUATIONS.  215 

(1)  The  sum  of  the  squares  of  two  consecutive  numbers  is 

481.     Find  the  numbers. 

Let  X  =  one  number, 

and  a;  +  1  =  the  other. 

Then             rr»  +  (x  +  1)2  =  481, 
or                    2ar^  +  2x  +  1  =  481. 
The  solution  of  which  gives,     x  =  15,  or  —  16. 
The  positive  root  15  gives  for  the  numbers,  15  and  16. 
The  negative  root  —  16  is  inapplicable  to  the  problem,  as  consecu- 
tive numbers  are  understood  to  be  integers  which  follow  one  another 
in  the  common  scale,  1,  2,  3,  4 

(2)  What  is  the  price  of  eggs  per  dozen  when  2  more  in  a 

shilling's  worth  lowers  the  price  1  penny  per  dozen  ? 
Let  X  =  number  of  eggs  for  a  shilling. 

Then,  -  =  cost  of  1  egg  in  shillings, 

X 

12 
and  —  =  cost  of  1  dozen  in  shillings. 

X 

But,  if  X  +  2  =  number  of  eggs  for  a  shilling, 

12 
=  cost  of  1  dozen  in  shillings. 

X  +  2 

12        I''         1 
.-. =i—  =  —  (1  penny  being  -j^^  of  a  shilling). 

X      X  +  2     12 
The  solution  of  which  gives  x  =  16,  or  — 18. 

And,  if  16  eggs  cost  a  shilling,  1  dozen  will  cost  |f  of  a  shilling, 
or  9  pence. 

Therefore,  the  price  of  the  eggs  is  9  pence  per  dozen. 

If  the  problem  be  changed  so  as  to  read :  What  is  the 
price  of  eggs  per  dozen  when  two  less  in  a  shilling's  worth 
raises  the  price  1  penny  per  dozen?  the  algebraic  statement 
will  be  12       12_1 

a;-2      X       12* 

The  solution  of  which  gives  x  =  18,  or  —  16. 

Hence,  the  number  18,  which  had  a  negative  sign  and  was  inappli- 
cable in  the  original  problem,  is  here  the  true  result. 


216  ALGEBRA. 


Exercise  XC. 


1.  The  sum  of  the  squares  of  three  consecutive  numbers  is 

365.     Find  the  numbers. 

2.  Three  times  the  product  of  two  consecutive  numbers 

exceeds  four  times  their  sum  by  8.    Find  the  numbers. 

3.  The  product  of  three  consecutive  numbers  is  equal  to 

three  times  the  middle  number.     Find  the  numbers. 

4.  A  boy  bought  a  number  of  apples  for  16  cents.     Had 

he  bought  4  more  for  the  same  money  he  would  have 
paid  i^  of  a  cent  less  for  each  apple.  How  many  did 
he  buy  ? 

5.  For  building  108  rods  of  stone-wall,  6  days  less  would 

have  been  required  if  3  rods  more  a  day  had  been 
built.     How  many  rods  a  day  were  built  ? 

6.  A  merchant  bought  some  pieces  of  silk  for  $  900.     Had 

he  bought  3  pieces  more  for  the  same  money  he  would 
have  paid  $  15  less  for  each  piece.  How  many  did 
he  buy  ? 

7.  A  merchant  bought  some  pieces  of  cloth  for  §168.75. 

He  sold  the  cloth  for  $  12  a  piece  and  gained  as  much 
as  1  piece  cost  him.  How  much  did  he  pay  for  each 
piece  ? 

8.  Find  the  price  of  eggs  per  score  when  10  more  in  62  J 

cents'  worth  lowers  the  price  31  i  cents  per  hundred. 

9.  The  area  of  a  square  may  be  doubled  by  increasing  its 

length  by  6  inches  and  its  breadth  by  4  inches.  De- 
termine its  side. 

10.  The  length  of  a  rectangular  field  exceeds  the  breadth 
by  1  yard,  and  the  area  is  3  acres.  Determine  its 
dimensions. 


QUADRATIC   EQUATIONS.  217 

11.  There  are  three  lines  of  which  two  are  each  ^  of  the 

third,  and  the  sum  of  the  squares  described  on  them 
is  equal  to  a  square  yard.  Determine  the  lengths  of 
the  lines  in  inches. 

12.  A  grass  plot  9  yards  long  and  6  yards  broad  has  a  path 

round  it.  The  area  of  the  path  is  equal  to  that  of 
the  plot.     Determine  the  width  of  the  path. 

13.  Find  the  radius  of  a  circle  the  area  of  which  would  be 

doubled  by  increasing  its  radius  by  1  inch. 

14.  Divide  a  line  20  inches  long  into  two  parts  so  that  the 

rectangle  contained  by  the  whole  and  one  part  may 
be  equal  to  the  square  on  the  other  part. 

15.  A  can  do  some  work  in  9  hours  less  time  than  B  can  do 

it,  and  together  they  can  do  it  in  20  hours.  How 
long  will  it  take  each  alone  to  do  it  ? 

16.  A  vessel  which  has  two  pipes  can  be  filled  in  2  hours 

less  time  by  one  than  by  the  other,  and  by  both  to- 
gether in  2  hours  55  minutes.  How  long  will  it  take 
each  pipe  alone  to  fill  the  vessel  ? 

17.  A  vessel  which  has  two  pipes  can  be  filled  in  2  hours 

less  time  by  one  than  by  the  other,  and  by  both  to- 
gether in  1  hour  52  minutes  30  seconds.  How  long 
will  it  take  each  pipe  alone  to  fill  the  vessel  ? 

18.  An  iron  bar  weighs^  36  pounds.     If  it  had  been  1  foot 

longer  each  foot  would  have  weighed  '}  a  pound  less. 
Find  the  length  and  the  weight  per  foot. 

19.  A  number  is  expressed  by  two  digits,  one  of  which  is  the 

square  of  the  other,  and  when  54  is  added  its  digits 
are  interchanged.     Find  the  number. 

20.  Divide  35  into  two  parts  so  that  the  sum  of  the  two 

fractions  formed  by  dividing  each  part  by  the  other 
may  be  2r^. 


218  ALGEBEA. 


21.  A  boat's  crew  row  3 J  miles  down  a  river  and  back 

again  in  1  hour  40  minutes.  If  the  current  of  the 
river  is  2  miles  per  hour,  determine  their  rate  of  row- 
ing in  still  water. 

22.  A  detachment  from  an  army  was  marching  in  regular 

column  with  5  men  more  in  depth  than  in  front.  On 
approaching  the  enemy  the  front  was  increased  by 
845  men,  and  the  whole  was  thu3  drawn  up  in  5  lines. 
Find  the  number  of  men. 

23.  A  jockey  sold  a  horse  for  §  144,  and  gained  as  much  per 

cent  as  the  horse  cost.     What  did  the  horse  cost  ? 

24.  A  merchant  expended  a  certain  sum  of  money  in  goods, 

which  he  sold  again  for  §  24,  and  lost  as  much  per 
cent  as  the  goods  cost  him.  How  much  did  he  pay 
for  the  goods  ? 

25.  A  broker  bought  a  number  of  bank  shares  (^100  each), 

when  they  were  at  a  certain  rate  per  cent  discount, 
for  $  7500 ;  and  afterwards  when  they  were  at  the 
same  rate  per  cent  premium,  sold  all  but  60  for  $5000. 
How  many  shares  did  he  buy,  and  at  what  price  ? 

26.  The  thickness  of  a  rectangular  solid  is  I  of  its  width, 

and  its  length  is  equal  to  the  sum  of  its  width  and 
thickness ;  also,  the  number  of  cubic  yards  in  its  vol- 
ume added  to  the  number  of  linear  yards  in  its  edges 
is  -J  of  the  number  of  square  yards  in  its  surface. 
Determine  its  dimensions. 

27.  If  a  carriage-wheel  16  i-  feet  round  took  1  second  more 

to  revolve,  the  rate  of  the  carriage  per  hour  would  be 
1^  miles  less.    At  what  rate  is  the  carriage  travelling? 


CHAPTER   XV. 

Simultaneous  Quadratic  Equations. 

237.  Quadratic  equations  involving  two  unknown  quan- 
tities require  different  methods  for  their  solution,  according 

to  the  form  of  the  equations. 

238.  Case  I.  When  from  one  of  the  equations  the  value 
of  one  of  the  unknown  quantities  can  be  found  in  terms  of 
the  other,  and  this  value  substituted  in  the  other  equation. 

Ex.   Solve:  3:^-2.y  =  5|  (1) 

x-y  =  2  i  (2) 

Transpose  x  in  (2),  y  =  x  —  2. 

Substitute  in  (1),  Sx^ -2x{x-2)  =  5. 

The  solution  of  which  gives  a;  =  1  or  —  5. 

.-.  y  =  -  1  or  —  7. 

Special  methods  often  give  more  elegant  solutions  of  examples  than 
the  general  method  by  substitution. 

I.  When  equations  have  the  form,  x  ±  y  =  a,  and  xy  =  b;  a^  ±  y^ 
=  a,  and  xy  =  b ;  or,  x  ±  y  =  a,  and  3^  +  y^  =  b. 

(1)   Solve:  "  +  ^='°|  P,! 

^  ^  a:y  =  300     J  (2) 

Square  (1),  x^-]-2xy+y^=  1600.  (3) 

Multiply  (2)  by  4,  4xy         =  1200.  (4) 

Subtract  (4)  from  (3),         x*  -  2  xy  +  y'  =  400. 
Extract  root  of  each  side.  a;  —  y  =  ±  20.  '     (6) 

Add  (1)  and  (G),  2a;  =  60or20, 

.-.a;  =  30  or  10. 
Subtract  (6)  from  (1),  2y  =  20  or  60, 

.•.y  =  10  or  30. 


220  ALGEBRA. 


^'  ^°i^«^       :^7/=4o} 


y  =  2or-6. 


x^y      20 
(3)    Solve:  ^  ^^        ^^ 


(1) 

(2) 


Square  (1),  rc2-2a;y +  y2=  16.  (3) 

Subtract  (2)  from  (3),  -  2  a;y  =  -  24.  (4) 

Subtract  (4)  from  (2),  ir^  +  2  xy  +  y^  ==  64. 

Extract  the  root,  a;  +  y  =  ±  8.  (5) 

By  combining  (5)  and  (1),  »  =  6  or  —2. 


(1) 
(2) 


Square  (1),  \^^^\  =  ^  (3) 

x^     xy     y^     400 

Subtract  (2)  from  (3),  1  =  1^.  (4) 

xy     400 

Subtract  (4)  from  (2),        \  -  A  +  \  =  ^. 
ar     xy     y^     400 

Extract  the  root,  =  +  —  (5) 

x     y      ~20  ^  ^ 

By  combining  (1)  and  (5),  cc  =  4  or  5. 

2/  =  5  or  4. 


6 


II.   WTien  one  equation  may  be  simplified  by  dividing  it  by  the  other. 
,^,    a  ,  ;r'  +  y'  =  91\  (1) 

Divide  (1)  by  (2),  a^  -xy  +  y''=  13.  (3) 

Square  (2),  x" +2xy  ^y^^Ad.  (4) 

Subtract  (3)  from  (4),  3  a;y  =  36. 

Divide  by  -  3,  -  rry  =  - 12.  (5) 

Add  (5)  and  (3),  x^-2xy +  y''=l. 

Extract  the  root,  x—y  =  ±l.  (6) 

By  combining  (6)  and  (2),  a;  =  4  or  3. 

y  =  3  or  4. 


SIMULTANEOUS   QUADRATIC   EQUATIONS. 


221 


Solve : 

Exercise  XCI. 

1.   a;  +  ?/=13| 
xy  =  m       J 

11. 

2.   a;  +  y  =  29| 
:ry=100     J 

12. 

:r3  +  y3  =  34n 
x  +  y  =  n      J 

3.   a:-y  =  19" 
a:y  =  66 

13. 

a;«  +  7/^  =  1008| 
a;  +  y=12        J 

4.   a;  —  ?/  =  45  ) 
.ry  =  250     J 

14. 

a.'«-2/«  =  98| 
a;~y  =  2      J 

5.   a:-y-:10      \ 

15. 

i^-y'=--279\ 
x-y=S        i 

6.   x-y  =  U      ' 

:,^ -1-^2  =  436  J 

16. 

a;-3y  =  l| 

a:y  +  y2  =  5/ 

7.   a;  +  y  =  12      | 

17. 

4y  =  5^'+l 
2a;y  =  33-:r2J 

:r      ?/      4 
a:*^y»      16  J 

la 

1-1=3      1 
1-1  =  21 

9.1  +  1  =  5      1 
X     y 

19. 

1-1  =  2}.    1 

X      y 

1           1          03 

10.    7x^-8xy  = 
52r  +  2y  =  7 

L59| 

20. 

a:2-2a:y-2/2 
2:  +  y-2 

^'} 

222  ALGEBRA, 

239.    Case  II.    When  each  of  the  two  equations  is  homo- 
geneous and  of  the  second  degree. 

Let  y  -^  vx,  and  substitute  vx  for  y  in  both  equations. 

From'(l),        2i^x'-4:v:t^  +  3x^  =  17, 


From  (2), 

Equate  the  values  of  x^, 

2v^-4:V  +  3      v'-V 

32t;2-64v  +  48=17i;2-17, 

15i;2-64v=-65. 


•.  x"^ 

17 

2v^- 

-4v 

+ 

3 

rV 

-x'  = 

16, 

■.x^  = 

16 

1* 

17 

16 

The  solution  gives, 

5        13 

Substitute  the  value  of  v  in 

x'-      1^ 

then, 

•.  a;=±3    or  ±f, 
3 

and 

y= 

K           13 

=  raj  =  db  5  or  ±7-. 
0 

c  T  Exercise  XCII. 

Solve : 

1.  x^  +  xi/-{-27/=74:    1  4.   .^2-42/2-9  =  01 
2rr2  +  2:ry  +  y2=73j  rry  +  2y2_3  =  0i 

2.  x'  +  xy  +  4:y'  =  6\  5.   :r2_^y_35^0| 


X'-XTZ  +  T/^^^I}  6.    a:2  4_^y_|_2y2  =  44 

2/^-2xy  =  -15    J  ^  22;2_^y_py2_16 


SIMULTANEOUS    QUADRATIC   EQUATIONS.  223 

7.   a^  +  xTZ-W^O)  9.    2ar^-}-^x^J-\-7/  =  70) 

_y2_2  =  0    J  C,x'  +  xy-i/  =  50    ) 


xi/  —  y^  —  2  =  0    J  (jx^-{-xy  —  y^ 

x^  —  XT/ -{- 9/^  =  7  \     10.   x^  —  xy  —  y^  =  i 

3:^+13:ry  +  8?/2  =  162J  2x^  +  ^xy  +  7f 


240.  Case  III.  When  the  two  equations  are  sym'metrical 
with  respect  to  x  and  y ;  that  is,  when  they  have  x  and  y 
similarly  involved  in  them. 

Thus,  the  equations  2r*  +  ?>a^f  +  2f,  2rry  -  3x  -  3y  +  1,  a;*- 
Zx^y  —  ?>  xif  +  3/*  are  symmetrical  equations. 

(1)    Solve:  -^y  =  18-y|  (1) 

^^  a;  +  y=12        J  (2) 

Fut  n  -^  V  for  X,  and  m  —  v  for  y,  in  (1)  and  (2). 

(1)  becomes        (u  +  vf  +  (u  —  v)'  =  18  (m  +  v)  (m  —  v), ' 

or  w3  +  3wt;*  =  9(w«-i;«).  (3) 

(2)  becomes  (m  +  r)  +  (m  —  v)  =  12, 
or  2w  =  12, 

:.u  =  6. 
Substitute  6  for  u  in  (3). 


(3)  becomes 
whence. 

216  +  18v«  =  9  (36  -  v*), 
r«  =  4. 

.•.r  =  ±2, 

and 

.-.  a;  =  w  +  V  =  6  ±  2  =  8  or  4, 
y  =  M-v  =  6T2  =  4or8. 

(2)    Solve: 

a;  +  y  =  8        1                                 (1) 
o;^ -1-2/^  =  706  J                                 (2) 

Put  M  +  i;  for  £C,  and  u  —  v-for  y,  in  (1)  and  (2). 

(1)  becomes  {yt,  +  v)  +  (w  —  v)  =  8, 

.*.  w  =  4. 

(2)  becomes  w*  +  eii'y^  ^  ^4  _  353^  (3^ 
Substitute  4  for  u  in  (3), 

256  +  96^2  +  v*  =  353, 
or,  r*  +  96r2  =  97.  (4) 

The  solution  of  (4)  gives  v  =  +  1  or  ±  V—  97. 

Taking  the  possible  values  of??,  x  —  5  or  3,  and  3/  =  3  or  5. 


224  ALGEBRA. 


!^ 


^qIyq  .  EXEECISE    XCIII. 

1.  4:Xi/  =  96~xY\  4.   4:(x  +  y)=-3xi/         ) 
x  +  7/  =  6            J  x  +  y  +  x'-~{-f  =  2Q) 

2.  ^2_|_^^13_^_y|  5^     4:X^  +  X7/  +  4:7/  =  bS    ^      -f 

XT/ =  6  J  bx'  +  5f  =  Q5  J 

3.  2(x'  +  f)  =  bx7j)  6.    :ry(:r  +  3/)-30| 
4  (:?;  —  3/)  =  :ry      J  x^-\-y^  =  S5        / 

241.  The  preceding  cases  are  general  methods  for  the  solution  of 
equations  which  belong  to  the  kinds  referred  to;  often,  however,  in 
the  solution  of  these  and  other  kinds  of  simultaneous  equations  in- 
volving quadratics,  a  little  ingenuity  will  suggest  some  step  by  which 
the  roots  may  easily  be  found. 

o  1     .  Exercise  XCIV. 

1.  37  —  y=7  I  8.    a;  —  3/=l      "I 
x'i-x7/  +  y'  =  lS}  x'  +  2/  =  8i} 

2.  :r2  +  .ry=35|  9.   x^  +  4:xy=S    ) 

3.  :ry-12  =  0|  10.    a;2-a:y +  3/2  =  48  I 
x-2y  =  ^  J                     .       a7-2/-8=:0        J 

4.  ^y_7  =  0    I  11.    a;2  +  3:?:y +3/2  =  1       | 
J  3a:2,         ,  3^_'l3/ 


2a;  — 53/ =  9        )  12.   :r2-2a:; 

a?  —  xy-\-y'^  =  1 }  x^  -\-xy 


5.   2a;-5y  =  9        )  12.   :^ -2xy -\-^y^  =\%\ 

7  J  :r2  +  rrv-v2  =  i.  J 


6.    ^  — 3/  =  9|  \'i.    X'\-y  =  a  1 

^1/  +  8  =  0  J  4:rv  -  a^  -  452  J 


^3/  +  8  =  OJ  ^xy 

7.    5a;  — 73/  =  0  'j      14.    a;  — 3/  =  ! 

5 


.a;  — 73/  =  0  ^      14.    a;  — 3/  =  !    ^ 

,^_13^  =  4-73/2  ^  +  ^  =  2* 

4  J  3/     a;  j 


SIMULTANEOUS    QUADRATIC   EQUATIONS. 


225 


15.    x^-{-9xy 
Ixy-f 


340  I 
171  i 


IQ.   x-\-y  =  Q>      I 

x?+f  =  n] 

17.  3:ry  +  2a;  +  2/  =  485) 
3a;-2y  =  0  J 

18.  a:  — y  =  l      I 

19.  a,-3  +  y'  =  2728         \ 
_.  ^-a;y +  2/^  =  124/ 

23.   a;  +  3/  =  «     \ 


21. 


22. 


23. 


^  +  y' 

^a^-Ax7j+5y'=9 

x  +  y  a:-y^lO 
a:-3/  a;  +  y  3 
0.^  +  2^  =  45 

1+1=5 

x     y 


17 


12 


.r+l     3/+I 

24.  a:2-a;y  +  /  =  7 

25.  a;  +  y  =  4      ) 


26.  x^-f 
x—y  = 

27.  a;^  —  ^y  =  a^  +  5^  I 
xy  —  y^^=2ab      i 

28.  ^-y2^4^j| 
rry  =  a^  —  P    J 

29.  a;y  =  0  1 

30.  x^  +  xy  +  y^  =  S7 

31.  ^  =  aa:  +  Z*?/  I 
y^  =  ay-\-bx) 

32.  a;  —  ?/  — 2  =  0 


33. 


34. 


x+y  x  —  y_m 
x  —  y  x-\-y  40 
6a:  =  20?/ +  9 


a;4  +  y4_82 

37.  3(a-l)-2(«  +  i)(3^_l) 
4(y  +  2)  =  i(a  +  2)         -^ 

38.  10a:2^i5^.y^3^j_2t,2>v 
10y2  +  15a;y  =  3a5-2^>2i 


^  +  ^=1 

^  +  ^  =  4 
a;      y 


^  +  y^  =  ^xy—  1/ 

36.   a.-^-y*  =  3093| 
a;-y  =  3         J 

-111 


22G  ALGEBRA. 


Exercise  XCV. 

1.  If  the  length  and  breadth  of  a  rectangle  were  each  in- 

creased by  1,  the  area  would  be  48  ;  if  they  were  each 
diminished  by  1,  the  area  would  be  24.  Find  the 
length  and  breadth. 

2.  The  sum  of  the  squares  of  the  two  digits  of  a  number  is 

25,  and  the  product  of  the  digits  is  12.  Find  the 
number. 

3.  The  sum,  the  product,  and  the  difference  of  the  squares, 

of  two  numbers  are  all  equal.     Find  the  numbers. 
Note.    Represent  the  numbers  hy  x  +  y  and  x—y,  respectively. 

4.  The  difference  of  two  numbers  is  f  of  the  greater,  and 

the  sum  of  their  squares  is  356.  What  are  the  num- 
bers? 

5.  The  numerator  and  denominator  of  one  fraction  are  each 
X4-1  ^17  greater  by  1  than  those  of  another,  and  the  sum  of 
"^^1   "    "r>    the  two  fractions  is  1  j\ ;  if  the  numerators  were  in- 

^  terchanged  the  sum  of  the  fractions  would  be  1|-, 

■r  —      "   .V  Find  the  fractions. 


6.  A  man  starts  from  the  foot  of  a  mountain  to  walk  to  its 
summit.  His  rate  of  walking  during  the  second  half 
of  the  distance  is  -^  mile  per  hour  less  than  his  rate 
during  the  first  half,  and  he  reaches  the  summit  in 
5i  hours.  He  descends  in  3f  hours,  by  walking  1 
mile  more  per  hour  than  during  the  first  half  of  the 
ascent.  Find  the  distance  to  the  top  and  the  rates 
of  walking. 
Note.  Let  2  re  =  the  distance,  and  y  miles  per  hour  =  the  rate  at  first. 
Then  -  +  -^—  =  5 J  hours,  and  -^  =  3|  hours. 

y    y-h  2/  +  1 


SIMULTANEOUS   QUADBATIO   EQUATIONS.  2127 

7.  The  sum  of  two  numbers  which  are  formed  by  the  same 

two  digits  in  reverse  order  is  ff  of  their  difference ; 
and  the  difference  of  the  squares  of  the  numbers  is 
3960.     Determine  the  numbers. 

8.  The  hypotenuse  of  a  right  triangle  is  20,  and  the  area 

of  the  triangle  is  96.     Determine  the  sides. 
Note.   The  square  on  the  hypotenuse  =  sum  of  the  squares  on  the 
sides ;  and  the  area  of  a  right  triangle  =  J  product  of  sides. 

9.  Two  boys  run  in  opposite  directions  round  a  rectangular 

field  the  area  of  which  is  an  acre ;  they  start  from 
one  corner  and  meet  13  yards  from  the  opposite  cor- 
ner ;  and  the  rate  of  one  is  ^  of  the  rate  of  the  other. 
Determine  the  dimensions  of  the  field. 

10.  A,  in  running  a  race  with  B,  to  a  post  and  back,  met 
him  10  yards  from  the  post.  To  make  it  a  dead  heat, 
B  must  have  increased  his  rate  from  this  point  41-f- 
yards  per  minute  ;  and  if,  without  changing  his  pace, 
he  had  turned  back  on  meeting  A,  he  would  have 
come  4  seconds  after  him.  How  far  was  it  to  the 
? 


IL  The  fore  wheel  of  a  carriage  turns  in  a  mile  132  times 
more  than  the  hind  wheel ;  but  if  the  circumferences 
were  each  increased  by  2  feet,  it  would  turn  only  88 
times  more.     Find  the  circumference  of  each. 

12.  A  person  has  $6500,  which  he  divides  into  two  parts 
and  loans  at  different  rates  of  interest,  so  that  the  two 
parts  produce  equal  returns.  If  the  first  part  had 
been  loaned  at  the  second  rate  of  interest,  it  would 
have  produced  $  180 ;  and  if  the  second  part  had  been 
loaned  at  the  first  rate  of  interest,  it  would  have  pro- 
duced $245.     Find  the  rates  of  interest. 


CHAPTER    XVI. 

Simple  Indeterminate  Equations. 

242.  If  a  single  equation  be  given  which  contains  two 
unknown  quantities,  and  no  other  condition  be  imposed, 
the  number  of  its  solutions  is  unlimited;  for,  if  any  value  be 
assigned  to  one  of  the  unknown  quantities,  a  corresponding 
value  may  be  found  for  the  other.  Such  an  equation  is 
said  to  be  indeterminate. 

243.  The  values  of  the  unknown  quantities  in  an  inde- 
terminate equation  are  dependent  upon  each  other ;  so  that, 
though  they  are  unlimited  in  number,  they  are  confined  to 
B,  particular  range. 

This  range  may  be  still  further  limited  by  requiring 
these  values  to  satisfy  some  given  condition ;  as,  for  instance, 
that  they  shall  be  positive  integers. 

244.  The  method  of  solving  an  indeterminate  equation 
in  positive  integers  is  as  follows : 

(1)    Solve  Zx-{-4iy  =  22,  in  positive  integers. 
Transpose,  3  a;  =  22  —  4  y, 

,.a,=  7-3/  +  ^, 
,         the  quotient  being  written  as  a  mixed  expression. 

.'.x  +  y-1      1-^ 


3 

Since  the  values  of  x  and  y  are  to  be  integral,  x  +  y  —  7  will  be 
integral,  and  hence,  "~  '^  will  be  integral,  though  written  in  the 
form  of  a  fraction. 

Let  ..  ~  V  =  m^  an  integer ; 


SIMPLE    INDETERMINATE    EQUATIONS.  229 


Then  1-3/  =  3m, 

.•.  3/  =  1  —  3m. 
Substitute  this  value  of  y  in  the  original  equation, 
3a;  +  4-  12m  =  22, 

:.x  =  Q  +  4  m. 
The  equation  y  =  \—om  shows  that  m  in  respect  to  y  may  be  0, 
or  have  any  negative  value,  but  cannot  have  a  positive  value. 

The  equation  a;  =  6  +  4m  shows  that  m  in  respect  to  x  may  be  0, 
but  cannot  have  a  negative  value  greater  than  1. 
.*.  m  may  be  0  or  —  1, 
and  then  a;  =  6,  3/  =  1, 

or  a;  =  2,  3/  =  4. 

(2)    Solve  5a;  —  14y  =  11,  in  positive  integers. 
Transpose,  5  a;  =  11  +  I43/, 

a;  =  2  +  23/  +  1+^. 

^  5      ' 

.•.  — - — ^  must  be  integral, 
o 

Now,  if      "t    -^  be  put  =  m,  then  y  =  "^^  ~    ,  a  fraction  in  form. 

0  4 

To  avoid   this  difficulty,   it  is  necessary  in  some  way  to  make 

the  coefficient  of  3/  equal  to  unity.     Since  — -"t— 2/  ig   integral,    any 

multiple  of  —^ — ^  is  integral.     Multiply,  then,  by  such  a  number  as 

5 
will  make  the  coefficient  of  y  greater  by  1  than  some  multiple  of  the 

denominator.     In  this  case,  multiply  by  4.     Then 

ii_162/  or  3y  +  i±i^  is  integral 
5  o 

4  +  V  •  . 

.'.  — —^  =  m,  an  integer ; 
5 
.'.  y  =  5m  —  4. 

Since       a;  =  |-  (11  +  14 3/),  from  the  original  equation, 

.•.  X  =  14m  —  9. 
Hero  it  is  obvious  that  m  may  have  any  positive  value,  and 

a;  =  5,  19,  33 

y  =  1,  6.  11 


230  ALGEBRA. 


The  required  multiplier  will  always  be  less  than  the  denominator ; 
and  for  this  reason  it  is  best  to  divide  the  original  equation  by  the 
smaller  of  the  two  coefficients,  in  order  to  have  the  multiplier  as 
small  as  possible. 

245.  The  necessity  for  a  multiplier  may  often  be  obviated 
by  a  little  ingenuity.     Thus, 

The  equation        4^/  =  29  —  7x      may  be  put  in  the  form  of 
4y  =  29  -8x  +  X, 
,:y  =  1-2x  +  ^-±^, 
in  which  the  fraction  is  of  the  required  form. 

The  equation         5  a;  =  18  +  13  y 
gives  x  =  3  +  2y+  ^^^-^, 

in  which  — i-2^  is  of  the  required  form, 
5 

246.  It  will  be  seen  from  (1)  and  (2)  that  when  only  pos- 
itive  integers  are  required,  the  number  of  solutions  will  be 
limited  or  unlimited  according  as  the  sign  connecting  x  and 
y  is  positive  or  negative. 

(3)    Find  the  least  number  that  when  divided  by  14  and  5 
will  give  remainders  1  and  3  respectively. 
If  iV  represent  the  number,  then 

N-l  aN-3 

=  X,  and =  y, 

14  '  5  ^' 

.-.  i\^=  14a;  +  1,  and  iV=  5?/  +  3, 
.-.  14a;  +  1  =  5?/  +  3. 
by  =  Ux  -  2, 
5y  =  15x  —  2  —  X, 

Q        2  +  a; 

,  .  y  =  6x . 

^  5 

2  +  X 
Let  — — —  =  m,  an  integer  ; 

5 
.•.  a;  =  5m  —  2. 

y  =  ^  (14a;  —  2),  from  original  equation, 
.•.  y  =  14m  —  6. 
If  m  =  1,  a;  =  3,  and  y  =  S, 

.-.  N=Ux  +  l  =  5y  +  3  ^  43.  Ans. 


SIMPLE   INDETERMINATE    EQUATIONS.  231 

(4)    Solve  5a:  +  6y  =  30,  so  that  x  may  be  a  multiple  of  y, 
and  both,  positive. 

Let  X  =  my. 

Then  (5^1  +  6)?/ =  30, 

30 


and  X  = 


5m  +  6 
30  m 


5m  +  6 

If  m  =  2,  X  =  3|,  y  =  If 

If  m  =  3,  jB  =  4f ,  y  =  If 

(5)    Solve  14  a;  +  22^/  =  71,  in  positive  integers. 

If  we  multiply  the  fraction  by  7  and  reduce, 
the  result  is  —  4y  -s-  J, 

a  form  which  shows  that  there  can  be  no  integral  solution. 

There  can  be  no  integral  solution  of  ax  ±  6y  =  c  if  a  and  b  have  a 
common  factor  not  common  also  to  c ;  for,  if  rf  be  a  factor  of  a  and 
also  of  b,  but  not  of  c,  the  equation  may  be  written, 

mdx  ±  ndy  =  c, 

or  mx  ±  ny  =  —,  a  fraction. 

a 

Exercise  XCVI. 
Solve  in  positive  integers  : 

1.  2:r  +  ll2/-49.  5.   3a;  +  8y  =  61. 

2.  7:r  +  3y==40.  6.   8a:  +  5y  =  97. 

3.  5a;  +  7y  =  53.  7.    16a:  + 7y  ==  110. 

4.  a: +102/ =  29.  ^    7a;  +  lOy  =  206. 
Solve  in  least  positive  integers : 

9.   12a;-72/  =  l.  12.   23a;-9y  =  929. 

10.  5a;-17y  =  23.  13.   23a;-33y  =  43. 

11.  23y-13a;-3.  14.   555a;-22y  =  73. 


232  ALGEBRA. 


15.  How  many  fractions  are  there  with  denominators  12 

and  18  whose  sum  is  -||-  ? 

16.  What  is  the  least  number  which,  when  divided  by  3 

and  5,  leaves  remainders  2  and  3  respectively  ? 

17.  A  person  counting  a  basket  of  eggs,  which  he  knows 

are  between  50  and  60,  finds  that  when  he  counts 
them  3  at  a  time  there  are  2  over ;  but  when  he 
counts  them  5  at  a  time  there  are  4  over.  How 
many  are  there  in  all  ? 

18.  A  person  bought  40  animals,  consisting  of  pigs,  geese, 

and  chickens,  for  $40.  The  pigs  cost  $5  a  piece,  the 
geese  $1,  and  the  chickens  25  cents  each.  Find  the 
number  he  bought  of  each. 

19.  Find  the  least  multiple  of  7  which,  when  divided  by  2, 

3,  4,  5,  6,  leaves  in  each  case  1  for  a  remainder. 

20.  In  how  many  ways  may  100  be  divided  into  two  parts, 

one  of  which  shall  be  a  multiple  of  7  and  the  other 
of  9? 

21.  Solve  18  a;  —  5?/  =  70  so  that  y  may  be  a  multiple  of  x, 

and  both  positive. 

22.  Solve  8a;  +  12y  =  23  so  that  re  and  y  may  be  positive, 

and  their  sum  an  integer. 

23.  Divide  70  into  three  parts  which  shall  be  multiples, 

respectively,  of  6,  7,  8. 

24.  Divide  200  into  parts  which  shall  give  integral  quo- 

tients when  divided  by  5,  7,  11,  and  the  sum  of  the 
quotients  shall  be  20. 

25.  A  number  consisting  of  three  digits,  of  which  the  mid- 

dle one  is  4,  has  the  digits  in  the  units  and  hundreds 
places  interchanged  by  adding  792.    Find  the  number. 


SIMPLE   INDETERMINATE   EQUATIONS.  233 

26.  Some  men  earning  each  $2.50  a  day,  and  some  women 

earning  each  $1.75  a  day,  receive  altogether  for  their 
daily  wages  $44.75.  Determine  the  number  of  men 
and  the  number  of  women. 

27.  A  wishes  to  pay  B  a  debt  of  £1  12  s.,  but  has  only  half- 

crowns  in  his  pocket,  while  B  has  only  4  penny -pieces. 
How  may  they  settle  the  matter  most  simply? 

28.  Show  that  323  a;  —  527y  =  1000  cannot  be  satisfied  by 

integral  values  of  x  and  y. 

29.  A  farmer  buys  oxen,  sheep,  and  hens.     The  whole  num- 

ber bought  is  100,  and  the  whole  price  £100.  If  the 
oxen  cost  £5,  the  sheep  £1,  and  the  hens  Is.  each, 
how  many  of  each  did  he  buy  ? 

30.  A  number  of  lengths  3  feet,  5  feet,  and  8  feet  are  cut ; 

how  may  48  of  them  be  taken  so  as  to  measure  175 
feet  all  together  ? 

31.  A  field  containing  an  integral  number  of  acres  less 

than  10  is  divided  into  8  lots  of  one  size,  and  7  of 
4  times  that  size,  and  has  also  a  road  passing  through 
it  containing  1300  square  yards.  Find  the  size  of 
the  lots  in  square  yards. 

32.  Two  wheels  are  to  be  made,  the '  circumference  of  one 

of  which  is  to  be  a  multiple  of  the  other.  What  cir- 
cumferences may  be  taken  so  that  when  the  first  has 
gone  round  three  times  and  the  other  five,  the  difier- 
ence  in  the  length  of  rope  coiled  on  them  may  be  17 
feet? 

33.  In  how  many  ways  can  a  person  pay  a  sum  of  £15  in 

half-crowns,  shillings,  and  sixpences,  so  that  the  num- 
ber of  shillings  and  sixpences  together  shall  be  equal 
to  the  number  of  half-crowns  ? 


CHAPTER    XVII. 

Inequalities. 

247.  Expressions  containing  any  given  letter  will  have 
their  values  changed  when  different  values  are  assigned  to 
that  letter ;  and  of  two  such  expressions,  one  may  be  for 
some  values  of  the  letter  larger  than  the  other,  for  other 
values  of  the  letter  smaller  than  the  other. 

Thus,  1  +  X  +  x^  will  be  greater  than  1  —  x  +  x^  for  all  positive 
values  of  x,  but  less  for  all  negative  values  of  x. 

248.  One  expression,  however,  may  be  so  related  to  an- 
other that,  whatever  values  may  be  given  to  the  letter,  it 
cannot  be  greater  than  the  other. 

Thus,  2x  cannot  be  greater  than  x^  +  1,  whatever  value  be  given 
to  X. 

249.  For  finding  whether  this  relation  holds  between 
two  expressions,  the  following  is  a  fundamental  proposition  : 

TT cc  and  b  are  unequal,  a^-\-b^>2ab. 

For,  {a  —  hf  must  be  positive,  whatever  the  values  of  a  and  b. 

That  is,  {a  -  bf  >  0, 

or  a'-2ab  +  b^>0; 

.'.  a^  +  h^>2  ab. 

250.  The  principles  applied  to  the  solution  of  equations 
may  be  applied  to  inequalities,  except  that  if  each  side  of 
an  •  equality  have  its  sign  changed,  the  inequality  will  be 
reversed. 

Thus,  if  a  >  6,  then  -  a  will  be  <  -  6. 


INEQUALITIES.  235 


(1)    If  a  and  h  be  positive,  show  that  c^  +  5^  is  >  c^h  +  aV^. 


if  (dividing  each  side  by  a  +  &), 


a^  -ah  ^V>  ah, 

if 

a^  +  h^>2ah. 

But 

a*  +  iMs  >  2  ah, 

.-.  a^  +  6^  >  a%  +  a6*. 

249. 


(2)    Show  th&t  a^  +  b^  +  c^  is  >ab+ac-\- he. 

Now,  a*  +  6'  is  >  2  aJ, 

a«  +  c»is>2ac,  ?  249. 

5»  +  c«  is  >  2  6c. 
By  adding,  2a'  +  2&«  +  2c*i8>2a6  +  2ac  +  26c, 
.'.  a'  +  6'  +  c*  is  >  a6  +  ac  +  he. 


Exercise  XCVII. 
Show  that,  the  letters  being  unequal  and  positive : 
1.   a^  +  3b'k>2b(a  +  b).  2.   a'b  + abHs>2a'b\ 

3.  (a^  +  P)  (a*  +  b*)  is  >  (a^  +  &')'• 

4.  a25  +  a2c  +  a52  +  52^  +  ac2  +  Z»c2is>6a5c. 

5.  The  sum  of  any  fraction  and  its  reciprocal  is  >  2. 

6.  lfa:^=a^~\-b^,  and  y^=c^-{-d^,xyis  >ac-{-bd,  or  ad-{-bo. 

7.  abc  is  >  (a  +  5  —  c)  (a  +  c  —  Z>)  (&  +  c  —  a). 

8.  Which  is  the  greater,  (a^  +  b^)  (^  +  c^')  or  (ac  +  Z>(^)2  ? 

9.  "Which  is  the  greater,  m^  -\-  7n  or  m^  -\- 1? 

10.  Which  is  the  greater,  a^ — Z>'*  or  4  a^  (a — b)  when  a  is  >  5  ? 

11.  Which  is  the  greater,  a/t"  +  'vI~  or  Va  +  V^? 

12.  Which  is  the  greater,  or  — ^  ? 

^  2  a  +  5 

13.  Which  is  the  greater,  — +  ^or-  +  -? 


CHAPTER  XVIII. 
Theory  of  Exponents. 

251.  The  expression  a",  when  n  is  a  positive  integer,  has 
been  defined  as  the  product  of  n  equal  factors  each  equal 
to  a.  §24. 

And  it  has  been  shown  that  a""  X  a""  =  a"*+".  §  66. 

That  a*"  -r-  a"*  =  a"*"",  if  m  be  greater  than  n;  §  93. 

or ,  if  m  be  less  than  n.  §  94. 

And  that  (a'")'*  =  a"*".  §  199. 

Also,  it  is  true  that  a""  X  h""  =  {aby ;  for 
(joiby  =  ah  taken  n  times  as  a  factor, 

=  a  taken  n  times  as  a  factor  X  h  taken  n  times  as  a 
factor  =  a"  X  S". 

252.  Likewise,  Va,  when  w  is  a  positive  integer,  has  been 
defined  as  one  of  tlie  n  equal  factors  of  a  (§  203)  ;  so  that  if 
^a  be  taken  n  times  as  a  factor,  the  resulting  product  is  a ; 
that  is,  {^aT  =  a.  

Again,  the  expression  Va"*  means  that  a  is  to  be  raised 
to  the  mth  power,  and  the  nth  root  of  the  result  obtained. 

And  the  expression  kVoT'  means  that  the  nth  root  of  a 
is  to  be  taken,  and  the  result  raised  to  the  mth  power. 

It  will  thus  be  seen  that  any  proposition  relating  to  roots 
and  powers  may  be  expressed  by  this  method  of  notation. 
It  is,  however,  found  convenient  to  adopt  another  method  of 
notation,  in  which  fractional  and  negative  exponents  are 
used. 


THEORY   OF   EXPONENTS.  237 

253.  The  meaning  of  a  fractional  exponent  is  at  once  sug- 
gested, by  observing  that  the  division  of  an  exponent,  when 
the  resulting  quotient  is  integral,  is  equivalent  to  extracting 
a  root.  Thus,  a^  is  the  square  root  of  a^,  and  3,  the  expo- 
nent of  a^  is  obtained  by  dividing  the  exponent  of  a^  by  2. 

If  this  division  be  indicated  only,  the  square  root  of  a* 
will  be  denoted  by  a^,  in  which  the  denominator  denotes 
the  root,  and  the  numerator  the  power.  If  the  same  mean- 
ing be  given  to  an  exponent  when  the  division  does  not  give 
an  integral  quotient,  a^  will  represent  the  square  root  of 
the  cube  of  a ;  and,  in  general,  a»,  the  nth  root  of  the  inth 
power  of  a.  This,  then,  is  the  meaning  that  will  be  as- 
signed to  a  fractional  exponent,  so  that  in  a  fractional 
exponent 

254.  The  numerator  will  indicate  a  power,  and  the  de- 
Twminator  a  root. 

255.  The  meaning  of  a  negative  exponent  is  suggested 
by  observing  that  in  a  series  of  descending  powers  of  a, 


a*,  a*,  a^,  a^,  a\ 


the  subtraction  of  1  from  the  exponent  is  equivalent  to  di- 
viding by  a ;  and  if  the  operation  be  continued,  the  result 


a",  a-\  a-\  «"»,  «"'» a"' 

Then,  aO_«_i      a-^  =  l-^a=^- 

a  a 

a-=^  =  --a  =  i;   a--  =  -. 

a  (J?  a" 


This,  then,  is  the  meaning  that  will  be  assigned  to  a  neg- 
ative exponent,  so  that, 

256.  A  mimher  with  a  negative  exponent  will  denote  the 
reciprocal  of  the  number  with  the  corresponding  positive 
exponent. 


238  ALGEBRA. 


It  may  be  easily  shown  that  the  laws  which  apply  to  pos- 
itive integral  exponents  apply  also  to  fractional  and  negative 
exponents. 

257.  To  show  that  <r  xV  =-  (abf  : 

a^  X  6^  =  Vo^  X  Vb^, 

=  (a5)»  (by  definition). 

Likewise  a"  X  Z>"  X  c"  =  {ahcf,  and  so  on. 

258.  To  show  that  (a-)«  =  a-« : 


Let 

x=-{a^y\ 

Then 

jc"  =  a»^,  and  a;'"'*  =  a. 

.-.  X  =  a^. 

But 

X  =r  (a^)"^         (by  supposition), 

.-.  (a^)^  =  a^. 

259.    To  show  that  O^  X  a""  =  a*""": 

Now 

am  X  a-«  =  a"^  X  — , 
a" 

=  ^  =  a^-n  if  w  >  n  : 
a" 

or 

=  -^  if  m  <  n, 

?93. 
§94. 


=  «-(«-"»)         (by  definition), 


260.    In  like  manner  the  same  laws  may  be  shown  to 
apply  in  every  case. 


THEORY    OF   EXPONENTS.  239 

261.    Hence,  whether  m  and  n  be  i7itegral  or  fractional, 
positive  or  negative : 

I.  a'"X  «'•  =  »'"+".  III.  («"')'*  =  a"'^ 

II.  cr  -f-  a"  =  a"*"**.  IV.  a"'Xh"'  =  {aby. 

Exercise  XCVIII. 
Express  with  fractional  exponents : 


2.  -vV?;  v^W;  va'^Z/V;  5V^?Z;^^ 

Express  with  radical  signs: 
3.    a^\  a^b^\  ^x^y~i]    3a;^y~^. 
Express  with  positive  exponents  : 

■"Write  in  the  form  of  integral  expressions : 

Simplify : 

6.  a^Xa^;  b^  X  h^  ;  c^  X  c^^ ;  d^  X  dj^. 

7.  m^  X  m~ ^ ;  n^  X  n~ tV  ;  a**  X  a^  ;  ft"  X  ft"^. 

8.  ft^  X  Vft ;  c"  ^  X  Vc  ;  ?/^  X  -^y ;  a;^  X  VaP. 

9.  ab^cXa'^hc^]  a^b^c'^  X  a^b'^c^d. 

10.   a;^2/^z^  X  x~^y~^z~^]  x^y^z^  X  x~^y~^z~^. 

n.„ix«-*xa-ix«-i;(f)^xg)*x(0 


240  ALGEBRA. 


12.  a^^a^;  c^ -^  c^~ ;  n^'^-^n^;  J  ~- -^a^ 

13.  (a«)*--(a«)t;  (c-i)f;  (m-iy;  (n*)-^  (xi)i 

14.  (p-^)-'^;  (^3)-i;  (a;-tyf)-t;  (at  x  af)-lf. 

15.  (4a-t)-f;  (27b-')- ^•,  (64c^«)-t;  (32d?-i«)i 

16.  ^i^^Y^  ^-M_V";  (3ta-r';  (Mir^ 

V8UV       VIC6-7  '^ 


262.  The  laws  that  apply  to  the  exponents  of  simple 
expressions  also  apply  to  the  exponents  of  compound  ex- 
pressions. 

(1)  Multiply  2/4  -j-yl  -j-  2/i  -f  1  by  2/4  —  1. 

2/1  +  2/3  +  2/1  +  1 

2/4-1 

y  +yl  +  yi  +  yh 

_  yl   -  y^  _  yl  _  1 

y  —  1  2/  —  1-  -4^s- 

(2)  Divide  x^  +  x^  -  12  by  a:*  -  3. 


/x^  -  3 
-12   -i 

^x^  +  4 


X'  +    a;* 


4a;*  -  12 

4x^  —  12  a;*  +  4.  Ans. 


,r  ,^.  T  Exercise  XOIX. 

Multiply : 

1.  x^^i-x^yP  +  y^^hyx^P  —  xPyP-^^/^p. 

2.  a;""*~"  —  2/"  by  x""  +  y»»~-«. 

3.  a;t-2;r^  +  l  byrr*-l. 

4.  8a'+4aU7  +  5a7Z*f  +  9Z'7by2a^-5i 


THEORY   OF   EXPONENTS.  241 

5.  l+ai-Ha'^"'byl-a&-^  +  a'Z'-l 

6.  a25-2  +  2  +  a-2Z.2bya2^-2-2-a-252. 

Divide  : 
a   a;*"  —  y^**  by  a;'*  —  y". 
9.   a^  +  y +  2  —  3^3^3  23  by  a;3+y^  + 2^- 

10.  x~\-i/  by  x^  —  x^y'^-^x^T/^  —  x^y^  +  y^- 

11.  a^y-^  +  2  +  x-Y^y^y~^  +  x'''y. 

12.  a-*  +  a-'  b-'  +  b'*hy  a''  -  a'"-  b'''  +  b-\ 

Find  the  squares  of: 

13.  4aZ'-^;  ah  —  b^;  a-\-a'^\  2aH^  —  a-^b^. 

If  a  =  4,  5  =  2,  c  =  1,  find  the  values  of: 

14.  aH;  bab-^)  2(aZ>)3  ;  oT^b-^c^;  \2a-^b-\ 

15.  Expand  (J- Z^a/;  (2a:-^  +  ^0*;  iab"^ -  by'^f. 

Extract  the  square  root  of: 

16.  9a;-*  -  18a;-3yi  +  15a;-2y  -  Qx-^y^-{-f. 

Extract  the  cube  root  of : 

17.  ^x^+  12a;2 _3o^_35_|. 45^-1  _^  21x-^ -  21x'\ 

Resolve  into  prime  factors  with  fractional  exponents : 
^18.    -v^r2,  -v/72,  v^96,  -v^ei ;  and  find  their  product. 

Simplify : 
19.    K-^-''")^X(.'r'^)-2J3^2.        20.    (a;^8«  X  ^-12)3^ 

21.  3  (J  -f  Z»2)2  _  4  (a^  5^)  (al  _  ^,1)  +  (ai  -  2  Z*^)l 

22.  \{crY-i\^-^\  24.    [{(a-'")-'*p]'--[J(a'")"i-^]-'. 


242  ALGEBRA. 

Eadical  Expressions. 

263.  An  indicated  root  ^hat  cannot  be  exactly  determined 
is  called  a  surd,  or  irrational  number.  An  indicated  root 
tbat  can  be  exactly  determined  is  said  to  have  the  form  of 
a  surd. 

264.  Surds  are  named  quadratic,  cubic,  biquadratic,  ac- 
cording as  the  second,  third,  or  fourth  roots  are  required. 

265.  The  product  of  a  rational  factor  and  a  surd  factor  is 
called  a  mixed  surd;  as,  3V2,  Z>V(X. 

266.  When  there  is  no  rational  factor  outside  of  the  rad- 
ical sign,  the  surd  is  said  to  be  entire;  as,  V2,  Va. 

267.  Since  Va  X  V^  X  V?=  ^abc,  the  products  of  two 
or  more  surds  of  the  same  order  will  be  a  radical  expression 
of  the  same  order  consisting  of  the  product  of  the  numbers 
under  the  radical  signs. 

268.  In  like  manner,  Vo^  =  Va^  X  V^=  aVb.    That  is, 
A  factor  under  the  radical  sign  whose  root  can  be  taken, 

may,  by  having  the  root  taken,  be  removed  from  under  the 
radical  sign. 

269.  Conversely,  since  asfb^^  Vc^, 

A  factor  outside  the  radical  sign  may  be  raised  to  the  cor- 
responding powe^^  and  placed  under  it. 


RADICAL   EXPRESSIONS. 


243 


270.  A  surd  is  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  integral  and  as  small  as  possible. 

271.  Surds  which,  when  reduced  to  the  simplest  form, 
have  the  same  surd  factor,  are  said  to  be  similar. 

Simplify : 

V50;     ■v'^IOS;     ^/l^' ]     ^\~\     ^ij^'^     -5^296352. 

(1)  V50=  V253<2  =  5\/2. 

(2)  v^l08  =  y/Wxl  =  3v^4. 


(3)    V7^'  =  <n¥fxf  =  yV7a^y\ 

,K,     4/  5a        4/40 aZ>c*       1    ^, 

(6)   v^296352. 


Hence,  296352  =  2^  x  3'  x  7=*, 

.-.  v/296352  =  y/7^  x  v^^  X  v^, 
=  7  X  3  X  2v^, 
=  42  v^.  Ans. 


23 

296352 

2« 

37044 

3» 

9261 

3 

1029 

7 

343 

7 

49 

In  simplifying  numerical  expressions  under  the  radical  sign,  the 
method  employed  in  (G)  may  be  used  with  advantage  when  the  factor 
whose  root  can  be  taken  is  not  readily  determined  by  inspection. 

Exercise  C. 
Express  as  entire  surds : 

1.  3V5;  3V2l;  5V32;  a^bWc;  xV^. 

2.  Zy^-sj3^y\  2x-Vxy',  a^-VoFb^;  Sc^-Vabc;  5abc-Vabc-^. 


3.  ^V^;  16VIII;  (x  +  y)^ 


xy 


^  +  2xy  +  y2 


244  ALGEBEA. 


Express  as  mixed  surds  : 


4.    V^;  V8a^;  VSl^^;  V24 ;  ^12^a^d^ 


5.    VlOOOa;  .V160^Y;  V108mV«;  Vl372^^ 


Simplify : 
7.   2-v^W^?^;  7V396^;  Q^^^ST^;  5V726. 
Q     aA  •    a/1  11  •    a/^  •    s-x/qrvs.   9-^/1  ' 

3l2^      3/^     ^^IA        f3a^ 

^Q     12  .        2      .  f^y^f  z'  \\:  fa%'\fc%W 


V5'  VlTOl'  \z'  )\xY)  ■'  V^-VV« 

11.  {ax)x(h^x)^;  (2a'b')  X  (b'x')^ ;  b(Sa%Sj)x{a'h-'f)i. 

12.  Show  that  V20,  V45,  Vf  are  similar  surds. 

1^ 

13.  Showtliat.2A/a'Z>2^  ^86^  i^y  are  similar  surds. 

14.  If   V2  =  1.414213,  find  the  values  of 

272.    Surds  of  the  same  order  may  be  compared  by  ex- 
pressing them  as  entire  surds. 

Ex.  Compare  |V7  and  fvTO. 

fVlO  =  V^. 

V^  =  V^,  and  V-?  =  VW. 

As  V-Vp-  is  greater  than  V^-,  |  VlO  is  greater  than  |V7. 


RADICAL   EXPRESSIONS.  245 

273.  The  product  or  quotient  of  two  surds  of  the  same 
order  may  be  obtained  by  taking  the  product  or  quotient 
of  the  rational  factors  and  the  surd  factors  separately. 

(1)  2  V5  X  5  V7  =  10  V35. 

(2)  9V5-i-3V7  =  8V|  =  3V||  =  fV35. 

Exercise  CI. 

1.  Which  is  the  greater  3  V7  or  2  Vl5? 

2.  Arrange  in  order  of  magnitude  9 V37  6V7,  SVlO. 

3.  Arrange  in  order  of  magnitude  4  V4,  3  V5,  5\^. 

4.  Multiply  3  V2  by  4  V6  ;  f  VlO  by  ^^^VIS. 

5.  Multiply  5  V|  by  f  Vl62 ;  i\/4  by  2-v^. 

6.  Divide  2 V5  by  3 VlS  ;  f  V2l  by  ^V^. 

7.  Simplify  fV3x  I  V5-^fV2. 

a   Simplify  2Vl0x^^-.±^. 
3V27     5Vli     15V21 

9.   Simplify  2-v^x5\/32-v-^yi08. 

274.  The  order  of  a  surd  may  be  changed  by  changing 
the  power  of  the  expression  under  the  radical  sign.      Thus, 

V5=^25;   ^=^. 

Conversely,      V25  =  V5  ;  V?  =  -Vc ; 

or,  in  general,  "V?*  =  "V^. 

In  this  way,  surds  of  different  orders  may  be  reduced  to 
the  same  order ^  and  may  then  be  compared,  multiplied,  or 
divided. 


246  ALGEBRA. 

(1)  To  compare  V2  and  V3. 

.•.  V3  is  greater  than  Vl. 

(2)  To  multiply  -y/Ta  by  VOx. 
-l^  =  (4a)i  =  (4a)o  =  v^i^^  _ 


16 


.-.  v^  X  V6^  =  -^^10^'  X  ^-"216^, 
=  \/16a2x216a^, 

=  v^26  X  2  X  S^^a^ar^, 
=  2y/5i^^.  Ans. 

(3)    To  divide  -^a  by  V63. 

.^  =  (3a)^  =  (3a)l  =  \/{^^  =  •^'^*; 
V66  =  (66)5  =  (6&)i  =  .^'pya  =  V2T6I?. 

.-.  ^3^-^  V6&=  ^9^2-^  ^&'216y3=  -^Cl^^ 
\24&3       \2=^x; 


'2453       \2=^x363' 


Wii^-^'^iss^'-^- 


Exercise  CII. 
Arrange  in  order  of  magnitude : 

1.  2-V/3,  3V2,  f-v^l.  3.   2^/22,  3^7,  4V2. 

2.  Vf,  ^iJ-  4.   3VI9,  5^2,  3^3. 


RADICAL   EXPEESSIONS.  247 

Simplify : 

5.  2 Vo^  X  -y/WcFb  X  V2bi  ;   ^/a^xy^  X  ^/oF^. 

6.  3(4a52)i  --  (2€^b)^  ;  (2a'P)^  X  (a*5^)-3  -  (a«Z.»)^.       >/ 

7.  (2  a5)^  X  (3  ab')^  -  (5  aJ^)*  ;  4 VT2  --  2  V3. 

9.   (7V2-5V6-3V8  +  4V20)x3V2. 
10.    V(||7xV(||)^;   ^/(4^*  X -v^'(2^. 

275.  In  tlie  addition  or  subtraction  of  surds,  each  surd 
must  be  reduced  to  its  simplest  form ;  and,  if  the  resulting 
surds  be  similar, 

Add  the  o'aiional  factors,  and  to  their  sum  annex  tJie  com- 
mon surd  factor. 

If  the  resulting  surds  be  not  similar, 
Connect  them  with  their  proper  signs. 

276.  Operations  with  surds  will  be  more  easily  performed 
if  the  arithmetical  numbers  contained  in  the  surds  be  ex- 
pressed in  their  prime  factors,  and  ii  fractional  exponents 
be  used  instead  of  radical  signs. 

(1)   Simplify  V27  +  V48  +  Vl47. 

Vis  =  (2*  X  3)^  =  2«  X  3^  =  4  X  3i  =  4\/3  ; 
Vli7  =  (7^  X  3)5  =  7  X  3^  =  7  V3~ 
.-.  \/27  +  \/48  +  ^/Wl  =  (3  +  4  +  7)\/3  =  14\/3.  Ans. 


248  ALGEBRA. 


(2)  Simplify  2\/320-3-v'40. 

2-\/S26  =  2  (2«  X  5)^  =  2  X  2^  X  5'  =  8^^; 
3  v^  =  3  (2'  X  5)^  =  3  X  2  X  5'  =  6  v^. 
.•.2v/320-3\^=8\/5-6v'5  =  2v^.  Ans. 

(3)  Find  the  square  root  of  Vsl. 

The  square  root  of  V^  =  (81»)^  =  81^  =  (S^)^ 
,^      =3l  =  (3-2)^=^/9. 

(4)  Find  the  cube  of  ^-v^.^ 

The  cube  of  |^=  (Jf  x  (2"6)3  _  i  x  2^  =  i  V2. 

Exercise  CIIL 
Simplify : 

1.  V27  +  2V48  +  3VI08;  3V1000  +  4V50  + 12V288. 

2.  ^/l2S+-\/6SE+^/lQ;  7-v/5i  +  3-v^+ -\/432. 

3.  12V72-3V128;  7a/8I  -  3^l029. 

4.  2V3  +  3Vi|-- V5l;  2V|+ V60- Vl5- V|. 

«•  Vf -V'iWS^^  3VI  +  2VS-4VS. 
6.    V4^+ V25a^>^-(a-52.)V^. 

8.  2a/40  + 3^/108+ -^500 --v^^20- 2 v'1372. 

9.  (2^/3^^)^  (3^3)2.  10.    (^Jft;    (V27)i 

11.  (-v/81)i;  (^5l2)i-;  (^256)-^  ^IG ;  ^27, 

12.  ^4;   V3"6;   ^32;   ^243 ;   -^^125 ;   ^4'9. 

13.  v^^  -J'^g^^;  -v/ie^;  -v/32^«. 


RADICAL   EXPRESSIONS.  249 

14.  (^8/;  (^27)^  (a/64/;  (■</4:)\ 

15.  (a-v/a)-^  {x^xyi;  (pWp)^  ]  (a'^A/O"^- 

Expand  by  tlie  method  explained  in  §  201 : 

16.  (Va+Vbf;  {^/m'+^^f;  (V^-2Vbf. 

17.  (2a2-iV^)«;  (2i^*-ify;  (?^-^J, 

-  (^-^)"  (£-#  e-0 

*(fNg-xlg"K"-'->'i(|-»'v^)' 

Find  the  square  root  of : 

21.  ^'^«  +  6a,'^"?/"  +  ll^'"y^**  +  6af'?/^'»  +  y<". 

22.  l  +  4a;-^-2:r-t-4^-^  +  25a;-^-24a;-t  +  16a;-l 

3 

277.  If  we  wish  to  find  the  approximate  value  of ,  it 

V2 
will  be  less  labor  to  multiply  first  both  numerator  and  de- 
nominator by  a  factor  that  will  render  the  denominator     , 
rational;  in  this  case  by  V2.     Thus,  )  "^ 

3   _      3V2      _3V2  "l  ' 

V2      V2xV2        2    * 

278.  It  is  easy  to  rationalize  the  denominator  of  a  frac- 
tion when  that  denominator  is  a  binomial  involving  only 
quadratic  surds.  The  factor  required  will  consist  of  the 
same  terms  as  the  given  denominator,  but  with  a  different 


250  ALGEBRA. 


sign  between  them.     Thus,  - — -—7=  will  have  its  denomi- 

d4-2V5 

nator  rationalized  by  multiplying  both  terms  of  the  fraction 

by6-2V5.     For, 

7  -  3  V5  _  (7  -  3  V5)  (6  -  2  V5) 
6  +  2  V5      (6  +  2  V5)  (6  -  2  V5) 

_72-32V5^9_2V5. 


16  2 

279.  By  two  operations  the  denominator  of  a  fraction 
may  be  rationalized  when  that  denominator  consists  of  three 
quadratic  surds. 

Thus,  if  the  denominator  be  V6  -\-  V3  —  ■\/2,  both  terms 
of  the  fraction  may  be  multiplied  by  V6  —  V3  +  V2.  The 
resulting  denominator  will  be  6  —  5  +  2V6  =  1  +  2  V6  ; 
and  if  both  terms  of  the  resulting  fraction  be  multiplied 
by  1  —  2  V6,  the  denominator  will  become  1  —  24  =  —  23. 

Exercise  CIV. 

Find  equivalent  fractions  with  rational  denominators,  for 
the  following : 

1  3  7  4- V2.         6 

V7  +  V5'  2V5-V6'  I  +  V2'  5-2V6* 

VJ— Vc     a—-\fh     ^xy  —  2y 
Find  the  approximate  values  of : 
3    A-  1  7V5         7  +  2VIO 

Vs'  V5-V2'  V74-V3'  7-2V10 


RADICAL   EXPRESSIONS. 


251 


Imaginary  Expressions. 
All  imaginary  square  roots  may  be  reduced  to  one 


form. 


V^^=  V:r2  X  (- 1)  =  W- 1. 
V^  =  Va  X  (—  1)  =  a^  V^. 

281.    V—  1  means  an  expression  which,  when  multiplied 
by  itself,  produces  —  1.     Therefore, 

(V- 1)» = ( V- 1/ X  V- 1 = - 1  v^ = -v=n: ; 

( V=^)'  =  (V^)»  X  (V=l)'  =  (- 1)  X  (- 1)  =  1 ; 

and  so  on.     So  that  the  successive  powers  of  V—  1  form  the 
repeating  series,  +V— 1,  —  1,  —  V—  1,  +  1. 

(1)  Multiply  1  +  V^  by  1  —  V^. 

1  +  V^=-l  +  2V~1: 
1  -  V^4  =  1  -  2V^1. 
(1  +  2V~1)(1  -  2\/="l)  =  1  -  4  (-  1)  =  5. 

(2)  Divide  V—  ab  by  V—  b. 

and  \/^6    =  b^V^l. 


V^b        bW-l 


Exercise  CV. 


Va. 


Multiply : 

1.  4  4-V^by4-V^;  V3-2V=^by  V3+2V=^. 

2.  V54  by  V—  2  ;  a  V—  b  by  a;  V—  y. 


252  ALGEBRA. 


4A5 


3.  V—  a  +  V—  h  by  V—  a  —  V— ^;  a  V—  a^^"*  by  V—  a^6^ 

4.  V=l0  by  V^  ;  2  V3  -  6  V^  by  4  VS  -  V^. 
Divide : 

5.  re V—  1  by  y V—  1 ;  1  by  V—  1 ;  a  by  a^V—  1. 


6.   V=l2by  V-3;  Vl5  by  V=^  ;  V=^by  V-20. 


Square  Root  of  a  Binomial  Surd. 

282.  The  product  or  quotient  of  two  dissimilar  quadratic 
surds  will  be  a  quadratic  surd.     Thus, 

Va&  X  ^s/abc  =  ah^/c; 
■y/abc  -^  Va6  =  -y/c. 

For  every  quadratic  surd,  when  simplified,  will  have  under  the 
radical  sign  one  or  more  factors  raised  only  to  the  first  power ;  and 
two  surds  which  are  dissimilar  cannot  have  all  these  factors  alike. 

Hence,  their  product  or  quotient  will  have  at  least  one  factor 
raised  only  to  the  first  power,  and  will  therefore  be  a  surd. 

283.  The  sum  or  difference  of  two  dissimilar  quadratic 
surds  cannot  he  a  rational  number,  nor  can  it  be  expressed 
as  a  single  surd. 

For  if  Va  ±  Vb  could  equal  a  rational  number  c,  we  should  have, 
by  squaring, 

a±2Vab  +  b  =  c^; 
that  is,  ±  2Vab  =  c'  —  a  —  b. 

Now,  as  the  right  side  of  this  equation  is  rational,  the  left  side 
would  be  rational ;  but,  by  §  282,  Vab  cannot  be  rational.  There- 
fore, Va  ±  Vb  cannot  be  rational. 

In  like  manner,  it  may  be  shown  that  Va  ±  Vb  cannot  be  ex- 
pressed as  a  single  surd  Vc. 


RADICAL    EXPEESSIONS.  253 

284.  A  quadratic  surd  cannot  equal  the  sum  of  a  rational 
number  and  a  surd. 

For,  if  Va  could  equal  c  +  V6,  we  should  have,  by  squaring, 
0==  c2  +  2cV6  +  6, 
and,  by  transposing,  2cVb  =  a  —  b  —  c^. 

That  is,  a  surd  equal  to  a  rational  number,  which  is  impossible. 

285.  If  a~\-  V6  =  x-{-  Vy,  then  a  will  equal  x  and  h 
will  equal  y. 

For,  by  transposing,  Vh  —  y/y  =  x  —  a;  and  if  b  were  not  equal 
to  2/,  the  difference  of  two  unequal  surds  would  be  rational,  which  by 
§  283  is  impossible. 

.'.  b  =  y  and  a  =  x. 

In  like  manner,  if  a  —  y/b  =  x  —  Vy,  a  will  equal  x  and  b  will 
equal  y. 

286.  To  extract  the  square  root  of  a  binomial  surd  a  +  V^. 


Let  Va  +  Vb  =  Vx+  Vy. 

Squaring,  a  +  Vb  =  x  +  2Vxy  +  y. 

.'.  x  +  y  =  a,  and  2Vxy  =-  Vb.  285. 

From  these  two  equations  the  values  of  a;  and  y  may  be  found. 

This  method  may  be  shortened  by  observing  that,  since 
Vb  =  2  V^,  _ 

a  —-y/b  =  x  —  2  V^y  +  y- 

By  taking  the  root,  v  «  _  Vb  =  Vx  —  Vy. 

...  (Va+V5)  (-Va-Vb)  =  ( Vi  +  Vy)  ( V^  -  Vy). 

.*.  Va^  —  b  =  x  —  i/. 

And,  as  a  =  x-]-7/, 

the  values  of  x  and  y  may  be  found  by  addition  and  sub- 
traction. 


254  ALGEBRA. 


(1)    Extract  the  square  root  of  7  +  4  V3. 

Let  v^  +  v^"  =  V?  +  4  \/3. 

Then  Vx  -  V^  =  V?  -  4  V3. 

By  multiplying,    x  —  y  =  V'49  —  48, 

.-.a: -2/  =  1. 
But  a;  +  2/  =  7, 

.•.  a;  =  4,  and  y  =  3. 

.-.  \/^+  Vy  =  2  +  V3. 


...  Vv  +  4  V3  -  2  +  Vs. 


Exercise  CVI. 


Extract  the  square  roots  of: 

1.  14  +  6V5.  6.   20-8V6.  11.  14-4V6. 

2.  17  +  4V15.         7.   9-6V2.  12.  38-12VI0. 

3.  IO  +  2V2I.         8.   94-42V5.  13.  103-12VT1. 

4.  I6  +  2V55.         9.    13-2V30.  14.  57-I2VI5. 

5.  9-2Vi4.  10.    II-6V2.  15.  SJ-VTO. 


16.  2a  +  2.-vV"^^l         18.   87  -  12 V42. 

17.  o?-2h-\/~c^^T\        19.   {a-\-hf-4.{a-h)Vab;        ^ 

287.  A  root  may  often  be  obtained  by  inspection.  For 
this  purpose,  write  the  given  expression  in  the  form  a+2  V^, 
and  determine  what  two  numbers  have  their  sum  equal  a, 
and  their  product  equal  h. 

(1)    Find  by  inspection  the  square  root  of  18  +  2 V77. 

It  is  required  to  find  two  numbers  whose  sum  is  18  and  whose 
product  is  77 ;  and  these  are  evidently  11  and  7. 
Then        18  +  2\/77  =  11  +  7  +  2V11  x  7, 

=  (Vil  +  V7)2. 
That  is,     Vn  +  V7=  square  root  of  18  +  2\/77. 


RADICAL    EXPRESSIONS.  255 

(2)    Find  by  inspection  the  square  root  of  75  —  12  V21. 

It  is  necessary  that  the  coefficient  of  the  surd  be  2  ;  therefore, 
75  —  12V2i  must  be  put  in  the  form  of 

75-2V756 
The  two  numbers  whose  sum  is  75  and  whose  product  is  756 
are  63  and  12. 
Then        75  -  2>/756  =  63  +  12  -  2\/63  x  12, 

=  (V63  -  Vl2)«. 
That  is,     V63  -  Vl2  =  square  root  of  75  -  12\/2i  ; 
or,  3\/7 -2\/3  =  8quarerootof  75  -  12\/2l. 

Equations  containing  Radicals. 

288.  An  equation  containing  a  single  radical  may  be 
solved  by  arranging  the  terms  so  as  to  have  the  radical  alone 
on  one  side,  and  then  raising  both  sides  to  a  power  corre- 
sponding to  the  order  of  the  radical. 

Ex.  V?^^  +  a;  =  9. 

V^"^  =  9  -  X. 
By  squaring,  x'  —  9  =  81  —  18  a;  +  ar*. 

18a:  =  90, 
.-.  a;  =  5. 

289.  If  two  radicals  be  involved,  two  steps  may  be 
necessary. 

Ex.  V^+l5+V5  =  15. 

Vx  +  15  +  \/x  =  15. 
By  squaring, 

a;  +  15  +  2  V?Tl5a;  +  a;  =  225. 


By  transposing,  2Var'  +  15a;  =  210  —  2a;. 

By  dividing  by  2,         Vx'^  +  15  a;  =  105  —  x. 
By  squaring,  x*  +  15x  =  11025  —  210  x  +  x*. 

225x  =  11025, 
.-.  x  =  49. 


256  ALGEBRA. 

Some  of  the  following  radical  equations  will  reduce  to 
simple  and  others  to  quadratic  equations. 

g  1  Exercise  CVII. 

1.  V:r  -  5  =  2.  6.    V^+4+V2^^--=6. 

2.  2VSx  +  ~4,-x  =  4:.  7.    Vl3^-1-V2rr-1  =  5. 

3.  3- V:^-l  =  2;r.  8.    V4+^  +  V^=^3. 


4.  V3:^;-2-2(:r-4).         9.    V25 +  a;+ V25-^=:  8. 

5.  4:X-12V^  =  16.  10.   a^  =  21  +  V^^^^. 


11.   2^;- V8:r3  +  26  +  2  =  0. 


12.  V^  +  1  +  V^  +  16  =  Vo;  4-  25. 

13.  ■V2x-j-l--Vx  +  4:  =  iVx^^.- 

14.  V^  +  3  +  V:r  4-  8  =  5  V^. 

15.    V3T^+V^=     ,  16.    V.^^=T+6^ 


V3  +  :^^  V:r2-r 

1^1  1 

17.      , — -^4- 


V^  +  1     Vx-l     V^-1 


18. 

V:r  +  2a 
V^  — 2  a 

+  Va 

20. 

;  — 2a_  :r 
;4-2a      2a 

3a; 
1    — 

+  V4:» 

--^^-2 

V7^4-44- 
V7^4-4- 

■2V3:i;- 

-1 

''   Zx 

-V4:r 

'-a;^        ■ 

-2V3a;- 

-1 

7, 


21.    ■\/{x  -  af  -\-  2ah  +  1)^  =  x  -  a  +  h. 


'   22.    V(a;  +  a)2  +  2a6  +  ^'  =  ^-a-a:. 


RADICAL   EXPRESSIONS.  257 

24.  Ax^~-^(x'^  +  l)(x^~2)=J(10-Sx^). 

25.  (a;!  -  2)  (xi'  -  4)  =  art  (:rt  -  1)'  -  12. 

26.  a;3_4^f:^96^  28.   a:* +  2^2^-^  =- 3  a. 

o-j 

27.  :i;  +  ^-'  =  2.9.  29.    81-v^  +  -|7^  =  52ar. 

s/x 

290.    Equations  may  be  solved  with  respect  to  an  expres- 
sio7i  in  the  same  manner  as  with  respect  to  a  letter. 

(1)    Solve  (r'-^r/- 8(^^-0;) +  12  =  0. 

Consider  (x*  —  x)  as  the  unknown  quantity. 
Then  (x«  -  a:)«  _  8  (x*  -  a:)  =  -  li 

Complete  the  square,  (a^  —  ar)*  —  ( )  +  16  =  4. 
Extract  the  root,  (x*  —  a:)  —  4  =  ±  2. 

a*  -  a;  =  6  or  2. 
Complete  the  square,  4  ar*  —  ( )  +  1  =  25  or  9. 

Extract  the  root,  2a;  —  1  =  ±5  or  ±3. 

2x=Q,  -4,4,  -2. 
.'.  aj=3, -2,  2, -1. 


(2)    Solve  5a:- 7:^2 -8V7ar^-5a;+ 1=8. 

Change  the  signs  and  annex  +  1  to  both  sides. 

7a.*  _ 5a;  +  1  +  8\/75»-5x  +  l  =  -  7. 
Solve  with  respect  to  VTa,-*  —  5x  +  1- 

(7x*  -  5a;  +  1)  +  8  (7a:«  -  5x  +  l)i  +  16  =  9. 
(7a;2_5x  +  l)^  +  4  =  ±3. 

(7x»_5a;  +  l)i  =  -lor-7. 
Square,  7a;*  —  5x  +  1  =  1  or  49. 

Transpose,  7ar*  —  5x  =  0  or  48. 

From  7a*  — 5x  =  0,        a;  =  Oorf; 
From  7a*-5x  =  48,      a;  =  3or-2f 

Note.  In  verifying  the  values  of  x  in  the  original  equation,  it  is 
seen  that  the  value  of  V7a;*  —  5a;  +  1  is  negative.  Thus,  by  putting 
0  for  X  the  equation  becomes  0  —  8Vl  =  8 ;  and  by  taking  —  1  for  VI 
we  have  (-  8)  (- 1)  =  8 ;  thatis,  8  =  8. 


258  ALGEBRA. 


(3)    Solve  x^  +  x+l-{-^  +  ~^^=l. 

X        XT 


("-:)■ 


Arrange  as  follows  :  (a;^  +  — |  +  (a;+-)  =  0. 

By  adding2to/'a'-^  +  i\ 

there  is  obtained  ar^  +  2  +  —  =  (  a;+-)  . 

x^      \         X I 

Multiply  by  4  and  complete  the  square, 

^%()  +  l  =  9. 

Extract  the  root,  2ja;  +  i}  +  l  =  ±3. 

2  fa; +  1^  =  2  or -4. 

a?  +  -  =  1  or  -  2. 

X 

Multiply  by  a,     a?  —  x  =  —  \,         and        x^  +  2x^  —  1. 
:Ax'-{)^l  =  -Z,  .-.^^2  + 2a; +  1  =  0. 

20^-1  =  +  ^/^  a; +  1  =  0. 

,\x=l{l±V^).  :.x==-l. 

291.  An  equation  like  that  of  (3)  which  will  remain  un- 
altered when  -  is  substituted  for  Xy  is  called  a  reciprocal 
equation. 

It  will  be  found  that  every  reciprocal  equation  of  odd  degree  will 
be  divisible  by  ar  —  1  or  a?  +  1  according  as  the  last  term  is  negative 
or  positive ;  and  every  reciprocal  equation  of  even  degree  with  its  last 
term  negative  will  be  divisible  by  a;''  —  1.  In  every  case  the  equation 
resulting  from  the  division  will  be  reciprocal. 


RADICAL   EXPRESSIONS.  .      259 

(4)    Solve^4-2x*-3:r3-3:r2  +  2a;  +  l  =  0. 

This  is  a  reciprocal  equation,  for,  if  x~^  be  put  for  x,  the  equation 
becomes  x-^  +  2x-'^ —  3x-^  —  3x-^  +  2x-'^  +  1  =  0,  which  multiplied 
by  a;* gives  1  +2x  —  3x^  —  3aP  +  2a^  +  a;^  =  0,  the  same  as  the  original 
equation. 

The  equation  may  be  written  {x^  +  1)  +  2x{s^  +  1)  —  3a^ (x  +  1)  =  0, 
which  is  obviously  divisible  by  a;  +  1.  The  result  from  dividing  by 
X  +  lis  a^  +  a^  —  4:0^  +  x  +  1  =  0,  or  {ai*  +  1)  +  x{a^  +  1)  =  ix^.  By  add- 
ing 2c^to{a^+l)  it  becomes  {x^  + 2x^  +  1)  =  {x^  +  If. 

Then  {x"  +  lf  +  x{z^  +  l)  =  Gx'. 

Multiply  by  4  and  complete  the  square, 

4(x2  +  l)2  +  (  )  +  x-2  =  25x«. 

Extract  the  root,  2{a^  +  l)+x  =  ±5x. 

Hence,  2  x*  +  2  =  4  a;  or  —  6  «. 

By  simplifying,  x*  — 2x  =  —  1 ;  and  a;*  +  3a;  =  — 1, 

whence,  x=l  and  1 ;     whence,  a;  =  ^  (—  3  ±  VE). 

Therefore,  including  the  root  —  1  obtained  from  the  factor 
a;  +  1,  the  five  roots  are  —  1,  1,  1,  ^  (—  3  ±  Vo). 

By  this  process  a  reciprocal  cubic  equation  may  be  reduced  to  a 
quadratic,  and  one  of  the  fifth  or  sixth  degree  to  a  biquadratic,  the 
solution  of  which  may  be  easily  effected. 


Solve  •  Exercise  CVIII. 


1.  3^-3x-6Vx^-Sx- 3  +  2  =  0. 

2.  ar^  +  Sx-^-{-\  =  ^. 

X        3? 

3.  (2a;2_3^.)2_  2(2:^-2 -3a:)  =  15. 


4.    {ax  —  J)^  +  4  a  {ax  —  h)  = 


'cr 


4 

5.  3{2a^-x)-{2x''-x)^  =  2. 

6.  lbx-2>a^-\-4:{x^-bx  +  bf^  =  lQ>. 

7.  a?  +  x-^-^x-\-x-^  =  ^.       9.    x^  +  x-\-\{x^  +  x)^  =  l. 

8.  a;2  4_  V?"^^  =  19.  10.    (.r  +  1)^  +  (:r  -  1)^  =  5. 


260  ALGEBRA. 


11.    :?;-l  =  2  +  2a:-^.  12.    V3r?;  +  5- V3:r-5  =  4. 

13.    (x'-{-l)-x(a^-\-l)  =  -2x'. 


14.    2a^-2-V2a^-bx  =  5(x+^). 


15.  x-\-2-4:X-\/x-\-2  =  l2f. 

16.  V2a;  +  a  +  V2a;-a==5. 


17.  V9x'  +  21x+l~-y/9x'  +  6x+l  =  Sx. 

18.  a;^_4a;t  +  a;-^  +  4a;-t=--J. 

19.  (2a7  +  3y)2-2(2a;+-33/)  =  81 
a:2-y2  =  21  J 

20.    a;  +  3/+ Va:  +  y  =  a"l  22.   xr^ -}- y^ -{- x -}- y.=  4:8 

x  —  y-\-  V^  —  y  =  b  J  xy^=\2 

r    21.   ^^*-:ry  +  y*  =  13|  23.    rr^  +  5;y-|- y^ ^  aH 

^  x^  —  xy  +  y^  =  ?t      )  ^  a;  +  ^xy  -\-y  =  h  ) 

24.    (ri:-y)2-3(rr-2/)  =  10| 


rr2y2_3a:y  =  54 

■\/x  —  Vy  =  a:^^  ( 
{£-^yY==2{x-yy 


25.    V^  —  Vy  =  a:^  (:r^  +  y^)  | 


26. 


a:-?/  —  (x  +  v")  =  54  -^ 


27.  a;  +  ?/+V^  =  28| 
^  +  y^  +  -'^2/  =  336  J 

28.  --3aa;=  V4^M^¥^  +  ^'. 
a  4 

29.  {x-\'l+x-^){x-l-{-x-^)  =  b\, 

30.  2(:r^-l)-i-2(a;^-4)-^  =  3(a;^-2)- 


VPv^A  (tV     OL     v^icX  r^  (^%/tc^^^^ 


^C^ 


/ 


CHAPTER  XIX. 
Logarithms. 

292.  In  the  common  system  of  notation  the  expression 
of  numbers  is  founded  on  their  relation  to  ten. 

Thus,  3854  indicates  that  this  number  contains  lO'  three  times,  10* 
eight  times,  10  five  times,  and  four  units. 

293.  In  this  system  a  number  is  represented  by  a  series 
of  different  powers  of  10,  the  exponent  of  each  power  being 
integral.  But,  by  employing /rac^ioria^  exponents,  any  num- 
ber may  be  represented  (approximately)  as  a  single  power 
of  10. 

294.  When  numbers  are  referred  in  this  way  to  10,  the 
exponents  of  the  powers  corresponding  to  them  are  called 
their  logarithms  to  the  base  10. 

For  brevity  the  word  "logarithm"  is  written  log. 


From  §  255  it  appears  that 


io»=    1,  io-M=A)    =1, 

10'=    10,  10-'(=T*Tr)   =01, 

10^=100,  10-»(=^^)  =  ,(X)1, 

and  so  on.     Hence, 

log      1=0,  log.l      =-1, 

log    10-1,  log  .01    =-2, 

log  100 -2,  log  .001 --3, 

and  so  on. 


262  ALGEBRA. 


It  is  evident  that  the  logarithms  of  all  numbers  between 

1  and      10  will  be       0  +  a  fraction, 

10  and    100  will  be       1  +  a  fraction, 

100  and  1000  will  be       2  +  a  fraction, 

1  and   .1      will  be  —  1  +  a  fraction, 

.1    and  .01    will  be  —  2  -}-  a  fraction, 

.01  and  .001  will  be  —  3  +  a  fraction. 

295.  The  fractional  part  of  a  logarithm  cannot  be  ex- 
pressed exactly  either  by  common  or  by  decimal  fractions  ; 
but  decimals  may  be  obtained  for  these  fractional  parts, 
true  to  as  many  places  as  may  be  desired. 

If,  for  instance,  the  logarithm  of  2  be  required ;  log  2  may  be  sup- 
posed to  be  \. 

Then  10'  =  2 ;  or,  by  raising  both  sides  to  the  ildrd  power,  10  =  8, 
a  result  which  shows  that  \  is  too  large.  ^ 

Suppose,  then,  log  2  =  j%.  Then  10^°  =  2,  or  by  raising  both  sides 
to  the  ienih  power,  10^  =  2^°.  That  is,  1000  =  1024,  a  result  which 
shows  that  j%  is  too  small. 

Since  \  is  too  large  and  -^^  too  small,  log  2  lies  between  \  and  -f^ ; 
that  is,  between  .33333  and  .30000. 

In  supposing  log  2  to  be  J,  the  error  of  the  result  is  ^^  =  x%  =  .2. 
In  supposing  log  2  to  be  -^^^  the  error  of  the  result  is  -^-^xWir^  "= 
f^y^  =  —  .024 ;  log  2,  therefore,  is  nearer  to  j%  than  to  \. 

The  difference  between  the  errors  is  .2  —  (—  .024)  ==  .224,  and  the 
difference  between  the  supposed  logarithms  is  .33333  —  .3  =  .03333. 

The  last  error,  therefore,  in  the  supposed  logarithm  may  be  consid- 
ered to  be  approximately  -^^j  of  .03333  ==  .0035  nearly,  and  this  added 
to  .3000  gives  .3035,  a  result  a  little  too  large. 

By  shorter  methods  of  higher  mathematics,  the  logarithm  of  2  is 
known  to  be  0.3010300,  true  to  the  seventh  place. 

296.  The  logarithm  of  a  number  consists  of  two  parts,  an 
integral  part  and  a  fractional  part. 

Thus,  log  2  =  0.30103,  in  which  the  integral  part  is  0,  and  the  frac- 
tional part  is  .30103 ;  log  20  =  1.30103,  in  which  the  integral  part  is  1, 
and  the  fractional  part  is  .30103, 


LOGARITHMS.  263 

297.  The  integral  part  of  a  logarithm  is  called  the  char- 
acteristic ;  and  the  fractional  part  is  called  the  mantissa. 

298.  The  mantissa  is  always  made  positive.  Hence,  in 
the  case  of  numbers  less  than  1  whose  logarithms  are  nega- 
tive, the  logarithm  is  made  to  consist  of  a  negative  charac- 
teristic and  a  positive  mantissa. 

299.  When  a  logarithm  consists  of  a  negative  character- 
istic and  a  positive  mantissa,  it  is  usual  to  write  the  minus 
sign  over  the  characteristic,  or  else  to  add  10  to  the  charac- 
teristic and  to  indicate  the  subtraction  of  10  from  the 
resulting  logarithm. 

Thus,  log  .2  =  1.30103,  and  this  may  be  written  9.30103  -  10. 

300.  The  characteristic  of  a  logarithm  of  an  integral  num- 
ber, or  of  a  mixed  number,  is  07ie  less  than  the  number  of 
integral  digits. 

Thus,  from  ^  294,  log  1  =  0,  log  10  =  1,  log  100  =  2.  Hence,  the 
logarithms  of  all  numbers  from  1  to  10  (that  is,  of  all  numbers  con- 
sisting of  one  integral  digit),  will  have  0  for  characteristic ;  and  the 
logarithms  of  all  numbers  from  10  to  100  (that  is,  of  all  numbers  con- 
sisting of  two  integral  digits),  will  have  1  for  characteristic ;  and  so 
on,  the  characteristic  increasing  by  1  for  each  increase  in  the  number 
of  digits,  and  therefore  always  being  1  less  than  tliat  number. 

301.  The  characteristic  of  a  logarithm  of  a  decimal  frac- 
tion is  negative,  and  is  equal  to  the  number  of  the  place  occu- 
pied by  the  first  significant  figure  of  the  decimal. 

Thus,  from  §  294,  log  .1  =  - 1,  log  .01  =  -2,  log  .001  =  -3.  Hence, 
the  logarithms  of  all  numbers  from  .1  to  1  will  have  —  1  for  a  charac- 
teristic (the  mantissa  being  plus) ;  the  logarithms  of  all  numbers  from 
.01  to  .1  will  have  —  2  for  a  characteristic ;  the  logarithms  of  all  num- 
bers from  .01  to  .001  will  have  —  3  for  a  characteristic ;  and  so  on, 
the  characteristic  always  being  negative  and  equal  to  the  number  of  the 
place  occupied  by  the  first  significant  figure  of  the  decim/iJ. 


264  ALGEBRA. 


302.  The  mantissa  of  a  logarithm  of  any  integral  num- 
ber or  decimal  fraction  depends  only  upon  the  digits  of  the 
number,  and  is  unchanged  so  long  as  the  sequence  of  the 
digits  remains  the  same. 

For,  changing  the  position  of  the  decimal  point  in  a  number  is 
equivalent  to  multiplying  or  dividing  the  number  by  a  power  of  10. 
Its  logarithm,  therefore,  will  be  increased  or  diminished  by  the  expo- 
nent of  that  power  of  10;  and,  since  this  exponent  is  integral,  the 
mantissa  of  the  logarithm  will  be  unaffected. 

Thus,  if  27196  =  lO^^^, 

then  2719.6  =  103-««, 

27.196  =  10^•«^^ 

2.7196  =  IQO-*^, 

.27196  =  I09«^-io, 

.0027196  =  lO''«»«-io. 

303.  The  advantage  of  using  the  number  10  as  tlie  base 
of  a  system  of  logarithms  consists  in  the  fact  that  the  man- 
tissa depends  only  on  the  sequence  of  digits,  and  the  charac- 
teristic on  the  position  of  the  decimal  point. 

304.  As  logarithms  are  simply  exponents  (§  294),  therefore, 
The  logarithm  of  a  p7vduct  is  the  sum  of  the  logarithms 

of  the  factors. 

Thus,  log  20  =  log  (2  X  10)  =  log  2  +  log  10 

=  0.3010  +  1.0000  =  1.3010 ; 
log  2000  =  log  (2  X  1000)  =  log  2  +  log  1000, 

=  0.3010  +  3.0000  =  3.3010 ; 
log  .2  =  log  (2  X  .1)  =  log  2  +  log  .1, 

=  0.3010  +  9.000  -  10  =  9.3010  -  10 ; 
log  .02  =  log  (2  X  .01)  =  log  2  +  log  .01, 

=  0.3010  +  8.0000  -  10  =  8.3010  -  10. 

Exercise  CIX. 

Given:  log  2  =  0.3010;  log  3  =  0.4771;  log  5  =  0.6990; 
log  7  =  0.8451. 

Find  the  logarithms  of  the  following  numbers  by  resolv- 


LOGARITHMS. 

265 

ing  the  numbers  into  factors,  and  taking 

the  sum  of  the 

logarithms 

of  the  factors : 

1.    log  6. 

9.   log  25. 

17.   log  .021. 

25. 

log  2.1. 

2.   log  15. 

10.    log  30. 

18.   log  .35. 

26. 

log  16. 

3.   log  21. 

11.   log  42. 

19.   log  .0035. 

27. 

log  .056. 

4.    log  14. 

12.   log  420. 

20.   log  .004. 

28. 

log  .63. 

5.   log  35. 

13.   log  12. 

21.    log  .05. 

29. 

log  1.75. 

6.   log  9. 

14.   log  60. 

22.    log  12.5. 

30. 

log  105. 

7.   log  8. 

15.    log  75. 

23.    log  1.25. 

31. 

log  .0105. 

8.   log  49. 

16.   log  7.5. 

24.    log  37.5. 

32. 

log  1.05. 

305.  As  logarithms  are  simply  exponents  (§  294),  there- 
fore, 

The  logaHthm  of  a  power  of  a  number  is  equal  to  the 
logarithm  of  the  numbe?'  multiplied  by  the  exponent  of  the 
power. 

Thus,  log  5'  =    7  X  log  5=    7x0.6990  =  4.8930. 

log  3"  =  11  X  log  3  =  11  X  0.4771  =  5.2481. 

306.  As  logarithms  are  simply  exponents  (§  294),  there- 
fore, when  roots  are  expressed  by  fractional  indices. 

The  logarithm  of  a  root  of  a  number  is  equal  to  the  loga- 
rithm of  the  number  m,ultiplied  by  the  index  of  the  root. 

Thus,  log  2*      =  J  oflog  2  =  J  X  0.3010  =  0.0753. 

log  .002*  =  J  of  (7.3010  -  10). 

The  expression  I  of  (7.3010  -  10) 

may  be  put  in  the  form  of  \  of  (27.3010  -  30)  which  =  9.1003  - 10 ; 
for,  since  20  -  20  =  0,  the  addition  of  20  to  the  7,  and  of  -  20  to  the 
—  10,  produces  no  change  in  the  value  of  the  logarithm. 

307.  In  simplifying  the  logarithm,  of  a  root  the  equal  pos- 
itive and  negative  numbers  to  be  added  to  the  logarithm  must 
be  such  that  the  resulting  negative  number,  when  divided  by 
the  index  of  the  root,  shall  give  a  quotient  of  —  10. 


266  ALGEBRA. 


Exercise  OX. 

Given:  log  2  =  0.3010;  log  3  =  0.4771 ;  log  5  =  0.6990 
log  7  =  0.8451. 

Find  logarithms  of  the  following  : 


1. 

2^ 

6.   5^ 

11. 

5i 

16.    7t. 

21. 

5i 

2. 

b\ 

7.    2i 

12. 

7tV. 

17.  5i 

22. 

2V. 

3. 

1\ 

8.    5i 

13. 

2l. 

18.    3tt. 

23. 

5t. 

4. 

3«. 

9.    3i 

14. 

5t. 

19.    7l. 

24. 

7-V-. 

5. 

7^ 

10.  7i 

15. 

3f. 

20.    3t. 

25. 

21^. 

308.  Since  logarithms  are  simply  exponents  (§  294),  there- 
fore, 

The  logarithm  of  a  quotient  is  the  logarithm  of  the  divi- 
dend minus  the  logarithm  of  the  divisor. 

Thus,      log  f  =  log  3  -  log  2  =  0.4771  -  0.3010  =     0.1761. 
log  I  =  log  2  -  log  3  =  0.3010  -  0.4771  -  -  0.1761. 

To  avoid  the  negative  logarithm  —0.1761,  we  subtract  the  entire 
logarithm  0.1761  from  10,  and  then  indicate  the  subtraction  of  10 
from  the  result. 

Thus,  -  0.1761  =  9.8239  -  10. 

Hence,  log  f  =  9.8239  -  10. 

309.  The  remainder  obtained  by  subtracting  the  loga- 
rithm of  a  number  from  10  is  called  the  cologarithm  of  the 
number,  or  arithmetical  complement  of  the  logarithm  of  the 
number. 

Cologarithm  is  usually  denoted  by  colog,  and  is  most 
easily  found  by  beginning  with  the  characteristic  of  the  loga- 
rithm and  subtracting  each  figure  from  9  down  to  the  last 
significant  figure,  and  subtracting  that  figure  from  10. 

Thus,  log  7  =  0.8451 ;  and  .colog  7  =  9.1549.  Colog  7  is  readily 
found  by  subtracting,  mentally,  0  from  9,  8  from  9,  4  from  9,  5  from 
9,  1  from  10,  and  writing  the  resulting  figure  at  each  step. 


LOGARITHMS.  267 


310.  Since  colog  7==  9.1549, 

and       log  f  =  log  1  -  log  7  =  0  -  0.8451  =  9.1549  -  10, 
it  is  evident  that, 

If  10  he  subtracted  from  the  cologaritlon  of  a  number^  the 
result  is  the  logarithm  of  the  reciprocal  of  that  nuinber. 

311.  Since  log  |  =  log  7  -  log  5, 

=  0.8451-0.6990  =  0.1461, 
and    log  7  +  colog  5  -  10  =  0.8451  +  9.3010  -  10, 

=  0.1461, 
it  is  evident  that, 

The  addition  of  a  cologarithm  —  10  is  equivalent  to  the 
subtraction  of  a  logarithm. 

The  steps  that  lead  to  this  result  are : 
l  =  7xi 
therefore,  •  log  1  =  log  (7  X  i)  =  log  7  +  log  \.  g  304. 

But  log  \  =  colog  5-10.  §  309. 

Hence,  log  |  =  log  7  +  colog  5  —  10. 

Therefore, 

312.  The  logarithm  of  a  quotient  w.ay  be  found  by  add- 
ing together  the  logaHthm  of  the  dividend  and  the  cologa- 
rithm of  the  divisor,  and  subtracting  10  from  the  result. 

In  finding  a  cologarithm  when  the  characteristic  of  the  logarithm 
is  a  negative  number,  it  must  be  observed  that  the  subtraction  of  a 
negative  number  is  equivalent  to  the  addition  of  an  equal  positive 
number. 

Thus.  log  -^  =  log  5  +  colog  .002  -  10, 

=  0.6990  +  12.6990  - 10, 
=  3.3980. 
Here  log  .002  =  3.3010,  and  in  subtracting  —  3  from  9  the  result  is 
the  same  as  adding  +  3  to  9. 
9 
Again,  log  ^^  =  log  2  +  colog  .07  —  10, 

=  0.3010  +  11.1549-10, 
=  1.4559. 


26^ 

I 

ALGEBRA. 

Also, 

log  --^  =  8.8451  -  10  +  9.0970  -  10, 
=  17.9421  -  20, 

. 

=  7.9421  - 10. 

Here, 

log  2=^  =  3  log  2  =  3  X  0.3010  =  0.9030. 

Hence, 

colog  2^  =  10  -  0.9030  =  9.0970. 

Exercise  CXI. 

Given  :  log  2  =  0.3010 ;  log  3  =  0.4771 ;  log  5  =  0.6990 : 
log  7  =  0.8451. 

Find  logarithms  for  tlie  following  quotients : 

1. 
2. 
3. 
4. 
5. 


313.  A  table  of  four-place  logarithms  is  here  given,  which 
contains  logarithms  of  all  numbers  under  1000,  the  decimal 
point  and  characteristic  being  omitted.  The  logarithms  of 
single  digits  1,  8,  etc.,  will  be  found  at  10,  80,  etc. 

Tables  containing  logarithms  of  more  places  can  be  pro- 
cured, but  this  table  will  serve  for  many  practical  uses,  and 
will  enable  the  student  to  use  tables  of  six-place,  seven- 
place,  and  ten-place  logarithms,  in  work  that  requires 
greater  accuracy. 


2 
5* 

7. 

5 
3' 

13. 

.05 
3' 

19. 

.05 
.003' 

25. 

.02*-' 
33  ■ 

2 

7" 

8. 

5 

2' 

14. 

.005 
2  * 

20. 

.007 
.02' 

26. 

3« 
.022' 

3 
5" 

9. 

7 
3* 

15. 

.07 
5  ' 

21. 

.02 
.007' 

27. 

73 

.022' 

3 

7" 

10. 

7 
2 

16. 

5 

.07' 

22. 

.005 
.07 

28. 

.073 
.0033" 

5 

7' 

11. 

3 

2' 

17. 

3 
.007 

23. 

.03 

7  * 

29. 

.005* 
73  • 

7 
5' 

12. 

7 
.5' 

18. 

.003 

7  * 

24. 

.0007 
.2 

30. 

73 
.0052' 

LOGARITHMS.  269 


314.  In  working  witli  a  four-place  table,  the  numbers 
corresponding  to  the  logarithms,  that  is,  the  antilogarithryis, 
as  they  are  called,  may  be  carried  io  four  dgnificant  digits. 

To  Find  the  Logarithm  of  a  Number  in  this  Table. 

315.  Suppose  it  is  required  to  find  the  logarithm  of  65.7. 
In  the  column  headed  "N"  look  for  the  first  two  significant 
figures,  and  at  the  top  of  the  table  for  the  third  significant 
figure.  In  the  line  with  65,  and  in  the  column  headed  7, 
is  seen  8176.  To  this  number  prefix  the  characteristic  and 
insert  the  decimal  point.     Thus, 

log  65.7  =  1.8176. 

Suppose  it  is  required  to  find  the  logarithm  of  20347. 
In  the  line  with  20,  and  in  the  column  headed  3,  is  seen 
3075 ;  also  in  the  line  with  20,  and  in  the  4  column,  is  seen 
3096,  and  the  difference  between  these  two  is  21.  The  dif- 
ference between  20300  and  20400  is  100,  and  the  difference 
between  20300  and  20347  is  47.  Hence,  yVo"  of  21  =  10, 
nearly,  must  be  added  to  3075.  That  is, 
log  20347  =  4.3085. 

Suppose  it  is  required  to  find  the  logarithm  of  .0005076. 
In  the  line  with  50,  and  in  the  7  column,  is  seen  7050;  in 
the  8  column,  7059 :  the  difference  is  9.  The  difference  be- 
tween 5070  and  5080  is  10,  and  the  difference  between  5070 
and  5076  is  6.  Hence,  ^  of  9  =  5  must  be  added  to  7050. 
That  IS,  ^^g  .0005076  =  6.7055  -  10. 

To  Find  a  Number  when  its  Logarithm  is  Given. 

316.  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  1.9736. 

Look  for  9736  in  the  table.  In  the  column  headed  "N," 
and  in  the  line  with  9736,  is  seen  94,  and  at  the  head  of 


270 


ALGEBRA. 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 
13 
14 

0000 
0414 
0792 
1139 
1461 

0043 
0453 
0828 
1173 
1492 

0086 
0492 
0864 
1206 
1523 

0128 
0531 
0899 
1239 
1553 

0170 
0569 
0934 
1271 
1584 

0212 
0607 
0969 
1303 
1614 

0253 
0645 
1004 
1335 
1644 

0294 
0682 
1038 
1367 
1673 

0334 
0719 
1072 
1399 
1703 

0374 
0755 
1106 
1430 
1732 

15 

16 
17 
18 
19 

1761 
2041 
2304 
2553 

2788 

1790 
2068 
2330 
2577 
2810 

1818 
2095 
2355 
2601 
2833 

1847 
2122 
2380 
2625 

2856 

1875 
2148 
2405 
2648 

2878 

1903 
2175 
2430 
2672 
2900 

1931 
2201 
2455 
2695 
2923 

1959 
2227 

2480 
2718 
2945 

1987 
2253 
2504 
2742 
2967 

2014 
2279 
2529 
2765 
2989 

20 

21 
22 
23 
24 

3010 
3222 
3424 
3617 
3802 

3032 
3243 
3444 
3636 
3820 

3054 
3263 
3464 
3655 
3838 

3075 
3284 
3483 
3674 
3856 

3096 
3304 
3502 
3692 
3874 

3118 
3324 
3522 
3711 

3892 

3139 
3345 
3541 

3729 
3909 

3160 
3365 
3560 
3747 

3927 

3181 
3385 
3579 
3766 
3945 

3201 
3404 
3598 
3784 
3962 

23 

26 
27 
28 
29 

3979 
4150 
4314 
4472 

4624 

3997 
4166 
4330 

4487 
4639 

4014 
4183 
4346 
4502 
4654 

4031 
4200 
4362 
4518 
4669 

4048 
4216 
4378 
4533 
4683 

4065 
4232 
4393 

4548 
4698 

4082 
4249 
4409 
4564 
4713 

4099 
4265 
4425 
4579 

4728 

4116 
4281 
4440 
4594 
4742 

4133 
4298 
4456 
4609 
4757 

30 

31 
32 
33 
34 

4771 
4914 
5051 
5185 
5315 

4786 
4928 
5065 
5198 
5328 

4800 
4942 
5079 
5211 
5340 

4814 
4955 
5092 
5224 
5353 

4829 
4969 
5105 
5237 
5366 

4843 
4983 
5119 
5250 

5378 

4857 
4997 
5132 
5263 
5391 

4871 
5011 
5145 
5276 
5403 

4886 
5024 
5159 
5289 
5416 

4900 
5038 
5172 
5302 
5428 

35 

36 
37 
38 
39 

5441 
5563 

5682 
5798 
5911 

5453 
5575 
5694 
5809 
5922 

5465 
5587 
5705 
5821 
5933 

5478 
5599 
5717 
5832 
5944 

5490 
5611 
5729 
5843 
5955 

5502 
5623 
5740 
5855 
5966 

5514 
5635 
5752 
5866 
5977 

5527 
5647 
5763 

5877 
5988 

5539 
5658 
5775 

5888 
5999 

5551 
5670 
5786 
5899 
6010 

40 

41 
42 
43 
44 

6021 
6128 
6232 
6335 
6435 

6031 
6138 
6243 
6345 
6444 

6042 
6149 
6253 
6355 
6454 

6053 
6160 
6263 
6365 
6464 

6064 
6170 
6274 
6375 
6474 

6571 
6665 
6758 
6848 
6937 

6075 
6180 
6284 
6385 
6484 

6085 
6191 
6294 
6395 
6493 

6096 
6201 
6304 
6405 
6503 

6107 
6212 
6314 
6415 
6513 

6117 
6222 
6325 
6425 

6522 

45 

46 
47 
48 
49 

6532 
6628 
6721 
6812 
6902 

6542 
6637 
6730 
6821 
6911 

6551 
6646 
6739 
6830 
6920 

6561 
6656 
6749 
6839 
6928 

6580 
6675 
6767 
6857 
6946 

6590 
6684 
6776 
6866 
6955 

6599 
6693 

6785 
6875 
6964 

6609 
6702 
6794 
6884 
6972 

6618 
6712 
6803 
6893 
6981 

50 

51 
52 
53 
54 

6990 
7076 
7160 
7243 
7324 

6998 
7084 
7168 
7251 
7332 

7007 
7093 
7177 
7259 
7340 

7016 
7101 

7185 
7267 
7348 

7024 
7110 
7193 
7275 
7356 

7033 
7118 
7202 

7284 
7364 

7042 
7126 
7210 
7292 

7372 

7050 
7135 

7218 
7300 
7380 

7059 
7143 

7226 
7308 
7388 

7067 
7152 
7235 
7316 
7396 

LOGARITHMS. 


271 


N 

0 

1 

2    3 

4 

5   6  j  7 

8 

9 

55 

56 
57 
58 
59 

7404 
7482 
7559 
7634 
7709 

7412 
7490 
7566 
7642 
7716 

7419 
7497 
7574 
7649 
7723 

7427 
7505 
7582 
7657 
7731 

7435 
7513 
7589 
7664 

7738 

7443 
7520 
7597 
7672 
7745 

7451 

7528 
7604 
7679 
7752 

7459 
7536 
7612 

7686 
7760 

7466 
7543 
7619 
7694 
7767 

7474 
7551 

7627 
7701 

7774 

60 

61 
62 
63 
64 

7782 
7853 
7924 
7993 
8062 

7789 
7860 
7931 
8000 
8069 

7796 

7868 
7938 
8007 
8075 

7803 
7875 
7945 
8014 
8082 

7810 
7882 
7952 
8021 
8089 

7818 
7889 
7959 
8028 
8096 

7825 
7896 
7966 
8035 
8102 

7832 
7903 
7973 
8041 
8109 

7839 
7910 
7980 
8048 
8116 

7846 
7917 
7987 
8055 
8122 

65 

66 
67 
68 
69 

8129 
8195 
8261 
8325 
8388 

8136 
8202 
8267 
8331 
8395 

8142 
8209 
8274 
8338 
8401 

8149 
8215 
8280 
8344 
8407 

8156 
8222 
8287 
8351 
8414 

8162 
8228 
8293 
8357 
8420 

8169 
8235 
8299 
8363 
8426 

8176 
8241 
8306 
8370 
8432 

8182 
8248 
8312 
8376 
8439 

8189 
8254 
8319 
8382 
8445 

70 

71 
72 
73 
74 

8451 
8513 
8573 
8633 
8692 

8457 
8519 
8579 
8639 
8698 

8463 
8525 
8585 
8645 
8704 

8470 
8531 
8591 
8651 
8710 

8476 
8537 
8597 
8657 
8716 

8482 
8543 
8603 
8663 

8722 

8488 
8549 
8609 
8669 

8727 

8494 
8555 
8615 
8675 
8733 

8500 
8561 
8621 
8681 
8739 

8506 
8567 
8627 
8686 
8745 

75 

76 
77 
78 
79 

8751 
8808 
8865 
8921 
8976 

8756 
8814 
8871 
8927 
8982 

8762 
8820 
8876 
8932 
8987 

8768 
8825 
8882 
8938 
8993 

8774 
8831 
8887 
8943 
8998 

8779 
8837 
8893 
8949 
9004 

8785 
8842 
8899 
8954 
9009 

8791 
8848 
8904 
8960 
9015 

8797 
8854 
8910 
8965 
9020 

8802 
8859 
8915 
8971 
9025 

80 

81 
82 
83 
84 

9031 
9085 
9138 
9191 
9243 

9036 
9090 
9143 
9196 
9248 

9042 
9096 
9149 
9201 
9253 

9047 
9101 
9154 
9206 
9258 

9053 
9106 
9159 
9212 
9263 

9058 
9112 
9165 
9217 
9269 

9063 
9117 
9170 
9222 
9274 

9069 
9122 
9175 
9227 
9279 

9074 
9128 
9180 
9232 

9284 

9079 
9133 
9186 
9238 
9289 

85 

86 
87 
88 
89 

9294 
9345 
9395 
9445 
9494 

9299 
9350 
9400 
9450 
9499 

9304 
9355 
9405 
9455 
9504 

9309 
9360 
9410 
9460 
9509 

9315 
9365 
9415 
9465 
9513 

9320 
9370 
9420 
9469 
9518 

9325 
9375 
9425 
9474 
9523 

9330 
9380 
9430 
9479 
9528 

9335 
9385 
9435 
9484 
9533 

9340 
9390 
9440 
9489 
9538 

90 

91 
92 
93 
94 

9542 
9590 
9638 
9685 
9731 

9547 
9595 
9643 
9689 
9736 

9552 
9600 
9647 
9694 
9741 

9557 
9605 
9652 
9699 
9745 

9562 
9609 
9657 
9703 
9750 

9566 
9614 
9661 
9708 
9754 

9571 
9619 
9666 
9713 
9759 

9576 
9624 
9671 
9717 
9763 

9581 
9628 
9675 
9722 
9768 

9586 
9633 
9680 
9727 
9773 

95 

96 
97 
98 
99 

9777 
9823 
9868 
9912 
9956 

9782 
9827 
9872 
9917 
9961 

9786 
9832 
9877 
9921 
9965 

9791 
9836 
9881 
9926 
9969 

9795 
9841 
9886 
9930 
9974 

9800 

9845 
9890 
9934 
9978 

9805 
9850 
9894 
9939 
9983 

9809 
9854 
9899 
9943 
9987 

9814 
9859 
9903 
9948 
9991 

9818 
9863 
9908 
9952 
9996 

272  ALGEBRA. 


the  column  in  which  973G  stands  is  seen  1.  Therefore, 
write  941,  and  insert  the  decimal  point  as  the  characteristic 
directs.     That  is,  the  number  required  is  94.1. 

Suppose  it  is  required  to  find  the  number  of  which  th,e 
logarithm  is  3.7936. 

Look  for  7936  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
7931  and  7938 ;  their  difference  is  7,  and  the  difference  be- 
tween 7931  and  7936  is  5.  Therefore,  -f-  of  the  difference 
between  the  numbers  corresponding  to  the  mantissas,  7931 
and  7938,  must  be  added  to  the  number  corresponding  to 
the  mantissa  7931. 

The  number  corresponding  to  the  mantissa  7938  is  6220. 

The  number  corresponding  to  the  mantissa  7931  is  6210. 

The  difference  between  these  numbers  is  10, 
and  6210 +  -f  of  10  =  6217. 

Therefore,  the  number  required  is  6217. 

Suppose  it  is  required  to  find  the  number  of  which  the 
logarithm  is  7.3882  -  10. 

Look  for  3882  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
3874  and  3892 ;  their  difference  is  18,  and  the  difference 
between  3874  and  3882  is  8.  Therefore,  -j^  of  the  differ- 
ence between  the  numbers  corresponding  to  the  mantissas, 
3874  and  3892,  must  be  added  to  the  number  corresponding 
to  the  mantissa  3874. 

The  number  corresponding  to  the  mantissa  3892  is  2450. 

The  number  corresponding  to  the  mantissa  3874  is  2440. 

The  difference  between  these  numbers  is  10, 
and  2440  + 1  of  10  =  2444. 

Therefore,  the  number  required  is  .002444. 


LOGAEITHMS.  273 


Exercise  CXII. 
Find  logarithms  of  the  following  numbers : 


1.  60. 

6.   3780.         Ul.    70633.             16.   877.08. 

2.   101. 

7.   54327.          12.    12028.             17.    73.896. 

3.   999. 

8.   90801.          13.   0.00987.          18.    7.0699. 

4.   9901. 

9.    10001.          14.   0.87701.           19.   0.0897. 

5.   5406. 

10.    10010.          15.    1.0001.             20.    99.778. 

Find  antilogarithms  to  the  following  logarithms : 

21.   4.2488. 

25.   4.7317.                 29.   9.0410-10. 

22.   3.6330. 

26.    1.9730.                 30.    9.8420-10. 

23.   2.5310. 

27.   9.8800-10.        31.    7.0216-10. 

24.    1.9484. 

28.   0.2787.                 32.    8.6580-10. 

Ex.   Find  the  product  of  908.4  X  .05392  X  2.117. 

log    908.4  =  2.9583 
log  .05392  =  8.7318 -10 
log    2.117  =  0.3257 

2.0158  =  log  103.7.  Ans. 

Find  by  logarithms  the  following  products : 

33.  948.76x0.043875.  35.  830.75x0.0003769. 

34.  3.4097x0.0087634.  36.   8.4395x0.98274. 

317.  When  any  of  the  factors  are  negative,  find  their  log- 
arithms without  regard  to  the  signs;  write  the  letter  n 
after  the  logarithm  that  corresponds  to  a  negative  number. 
If  the  number  of  logarithms  so  marked  be  odd,  the  product 
is  negative ;  if  even,  the  product  is  positive. 


274 


ALGEBEA. 


Find  the  products  of : 

37.  7564 X  (-0.003764). 

38.  3.7648  X  (-0.083497). 

Ex.  Find  the  quotient  of 


39.  -5.840359 X  (-0.00178). 

40.  -8945.07x73.846. 


8.3709x834.637 


log  8.3709 
log  834.637 
colog  7308.946 


Find  the  quotients  of : 
70654 


41. 


42. 


43. 


44. 


45. 


54013' 

58706 
93078' 

8.32165 
0.07891* 

65039 
90761* 

7.652 
-0.06875* 


46. 


47. 


48. 


7308.946 

0.9227 

2.9215 

6.1362-10 

9.9804 -10  =  log  .9558.  Ans. 


0.07654 


83.947  X  0.8395 

7564  X  0.07643 
8093  X  0.09817' 

89  X  753  X  0.0097 
36709  X  0.08497  * 

413  X  8.17  X  3182 
915  X  728  X  2.315' 


212  x(- 6.12)  x(- 2008) 
*       365  X  (- 531)  X  2.576    * 


Ex.  Find  the  cube  of  .0497. 

log  .0497  =  8.6964  - 
3 

10 

00012: 

6.0892- 

■  10  =  log  . 

28.  Ans. 

Find  by  logarithms : 

51.    6.05^. 

55.    0.78765^ 

59. 

(10|)^ 

63. 

(mr'- 

52.    1.051^ 

56.    0.691^ 

60. 

my- 

64. 

a^r- 

53.    1.1768^ 

57.  (ny\ 

61. 

(mr- 

65. 

m'-'- 

54.    1.3178^«. 

58.    (il)^ 

62. 

(7_«_)o.3a 

.    66. 

myr^'. 

LOGARITHMS.  275 


Ex.    Find  the  fourth  root  of  0.00862. 

log  0.00862=    7.9355-10 
30.         -  30 
4)37.9355-40 
•  9.4839  -  10  =  log  .3047.  Ans. 
Find  by  logarithms : 

67.  7*.  70.  8379^.  73.  0.17643i     76.  {-f^i^)^. 

68.  11^.  71.  906.80i  74.  2.5637".     77.  (9||)i 

69.  783i         72.  8.1904i  75.  (ffi)l         78.  (11-^)^ 

Find  by  logarithms  the  values  of : 

10.00754332  X  78.343  X  8172.4^  x  0.00052 


79. 
80. 
81. 
82. 
83. 
84. 
85. 
86. 


i 


64285.*  X  154.27*  X  0.001  X  586.79^ 
15.832^  X  5793.6*  X  0.78426 


\  0.000327*  X  768.94«X  3015.3  x0.007i' 

J7.1895  X  4764.2^  X  0.00326^ 
Sf  0.00048953  X  457^  X  5764.4^" 

5|3.1416x  4771.21  X2.7183i 

^30.: 

J732.( 

\42.2'] 

\        0.03274  X  0.6428       ' 
3/    7.1206  X  V0.13274  X  0.057389 


.103*  X  0.4343i  X  69.897* 

/0.032712  X  53.429  X  0.77542« 
32.769  X  0.000371* 

^732.056'  X  0.0003572*  X  89793 
.2798*  X  3.4574  X  0.0026518* 

(7932x0.00657x0.80464 


VO.43468  X  17.385  X  VO.0096372 

^      r  3.075526^  X  5771.2^  x  0.0036984^  x  7.74  ]^ 
'     I  72258  X  327.933  X  86.97*  J* 


276^'  ALGEBRA. 


318.  Since  any  positive  number  other  than  1  may  be 
taken  as  the  base  of  a  system  of  logarithms,  the  following 
general  proofs  to  the  base  a  should  be  noticed. 

I.  The  logarithm  of  the  product  of  two  or  raore  numbers 
is  equal  to  the  sum  of  the  logarithvis  of  the  numbers. 

For,  let  m  and  n  be  two  numbers,  and  a;  and  y  their  logarithms. 
Then,  by  the  definition  of  a  logarithm,  m  ==  a==,  and  n  --=  av. 
Hence,  m  x  n  =  a'' y.  cO>  =  a^'^v. 

.'.  log  (mxn)  =  x  +  y, 

=  log  m  +  log  n. 

In  like  manner,  the  proposition  may  be  extended  to  any  number 
of  factors. 

II.  The  logarithm  of  a  quotient  is  equal  to  the  logarithm, 
of  the  dividend  Tniniis  the  logarithm  of  the  divisor. 

For,  let  m  and  n  be  two  numbers,  and  x  and  y  their  logarithms. 
Then  m  =  a*  and  n  =  av. 

Hence,  m  -=-  n  =  a*  h-  a^  =  a^-y. 

.*.  log  {m  ^n)  =  x  —  y, 

=  log  m  —  log  n. 

From  this  it  follows  that  log  —  ==  log  1  —  log  m. 
But,  since  log  1  =  0,  log  —  =  —  log  m. 

III.  The  logarithm  of  a  pjbwer  of  a  number  is  equal  to 
the  logarithm  of  the  number  m^uUiplied  by  the  exponent  of 
the  power. 

For,  let  X  be  the  logarithm  of  m. 

Then  m  =  a", 

and  mP  ^  {w^y  =  o^. 

.*.  log  mP  ^  px, 

=  p  log  m. 


LOGARITHMS.  277 


IV.    The  logarithm  of  the  root  of  a  number  is  equal  to  the 
logarithm  of  the  numher  divided  hy  the  index  of  the  root. 

For,  let  X  be  the  logarithm  of  m. 


Then  m  =  a' 

i      X      log  m 


and  m*  =  {a^f  =  S 


:.  log 


m' 


319.  An  exponential  equation,    that   is,  an   equation   in 

which  the   exponent  is   the   unknown    quantity,  is  easily- 
solved  by  logarithms. 

For,  let  a*  =  m. 

Then  log  a*  =  log  m, 

.'.  X  log  a  =  log  m, 

•  a;  -  ^Qg  ^ 
log  a" 

Ex.  Find  the  value  of  x  in  81*  =  10. 
81*  =10, 

^_  log  10 
log  81' 
.'.  log  X  =  log  log  10  +  colog  log  81, 
=  0  +  9.7193  -  10, 
.•.»  =  . 524. 

320.  Logarithms  of  numbers  to  any  base  a  may  be  con- 
verted into  logarithms  to  any  other  base  h  by  dividing  the 
computed  logarithms  by  the  logarithm  of  b  to  the  base  a. 

For,  let  log  m  =  y  to  the  base  b, 

and  log   h  =  X  to  the  base  a. 

Then  m  =  by,  and  b  =  a", 

.-.  m  =  (a*)y  =  a'y. 

.'.  log  m  (to  base  a)  =  ccy  =  log  b  (to  base  a)  X  log  ?n  (to  base  &), 

1  /,    1         IN      log  m  (to  base  a) 

.•.logm(tobase6)  =  ^;|^A___i. 

This  is  usually  written,  log6  m  =  ,'"''  ^. 

log«6 


CHAPTER  XX. 

Ratio,  Proportion,  and  Variation. 

321.  The  relative  magnitude  of  two  numbers  is  called 
their  ratio,  and  is  expressed  by  the  fraction  which  the  first 
is  of  the  second. 

Thus,  the  ratio  of  6  to  3  is  indicated  by  the  fraction  |,  which  is 
sometimes  written  6 :  3. 

322.  The  first  term  of  a  ratio  is  called  the  antecedent, 
and  the  second  term  the  consequent.  When  the  antecedent 
is  equal  to  the  consequent,  the  ratio  is  called  a  ratio  of 
equality ;  when  the  antecedent  is  greater  than  the  conse- 
quent, the  ratio  is  called  a  ratio  of  greater  inequality ;  when 
less,  a  ratio  of  less  inequality. 

323.  When  the  antecedent  and  consequent  are  inter- 
changed, the  resulting  ratio  is  called  the  inverse  of  the  given 
ratio. 

Thus,  the  ratio  3  :  6  is  the. inverse  of  the  ratio  6 :  3. 

324.  The  ratio  of  two  quantities  that  can  be  expressed  in 
integers  in  terms  of  a  common  unit  is  equal  to  the  ratio  of 
the  two  numbers  by  which  they  are  expressed. 

Thus,  the  ratio  of  $9  to  $11  is  equal  to  the  ratio  of  9  :  11 ;  and  the 
ratio  of  a  line  2f  inches  long  to  a  line  3  J  inches  long,  when  both  are 
expressed  in  terms  of  a  unit  -^^  of  an  inch  long,  is  equal  to  the  ratio 
of  32  to  45. 

325.  Two  quantities  different  in  hind  can  have  no  ratio, 
for  then  one  cannot  be  a  fraction  of  the  other. 


RATIO.  279 


326.  Two  quantities  that  can  be  expressed  in  integers  in 
terms  of  a  common  unit  are  said  to  be  commensurahle. 
The  common  unit  is  called  a  common  m,easure,  and  each 
quantity  is  called  a  multiple  of  this  common  measure. 

Thus,  a  common  measure  of  2^  feet  and  3f  feet  is  ^  of  a  foot,  which 
is  contained  15  times  in  2\  feet,  and  22  times  in  3|  feet.  Hence,  2\ 
feet  and  3f  feet  are  multiples  of  ^  of  a  foot,  2\  feet  being  obtained 
by  taking  ^  of  a  foot  15  times,  and  3f  by  taking  ^  of  a  foot  22  times. 

327.  When  two  quantities  are  incommensurable,  that  is, 
have  no  common  unit  in  terms  of  which  both  quantities  can 
be  expressed  in  integers,  it  is  impossible  to  find  a  fraction 
that  will  indicate  the  exact  value  of  the  ratio  of  the  given 
quantities.  It  is  possible,  however,  by  taking  the  unit  suf- 
ficiently small,  to  find  a  fraction  that  shall  difier  from  the 
true  value  of  the  ratio  by  as  little  as  we  please. 

Thus,  if  a  and  h  denote  the  diagonal  and  side  of  a  square, 

•J=V2. 
o 

Now  V2  =  1.41421356 a  value  greater  than  1.414213,  but  less 

than  1.414214. 

If,  then,  a  millionth  part  of  h  be  taken  as  the  unit,  the  value  of  the 
ratio  I  lies  between  jHMM  ^^^  xt^MM-  ^.nd  therefore  differs  from 
either  of  those  fractions  by  less  than  iJSJS'h'Uis^- 

By  carrying  the  decimal  farther,  a  fraction  may  be  found  that  will 
differ  from  the  true  value  of  the  ratio  by  less  than  a  billionth,  tril- 
lionth,  or  any  other  assir/ned  value  whatever. 

328.  Expressed  generally,  when  a  and  b  are  incommen- 
surable, and  b  is  divided  into  any  integral  number  (n)  of 
equal  parts,  if  one  of  these  parts  be  contained  in  a  more 
than  771  times,  but  less  than  m  + 1  times,  then 

on  n 

that  is,  the-  value  of  -  lies  Between  —  and  — ^t_. 
h  n  n 


280  ALGEBRA. 

The  error,  therefore,  in  taking  either  of  these  values  for 

-  is  <  -.  But  by  increasing  n  indefinitely,  -  can  be  made 
on  n 

to  decrease  indefinitely,  and  to  become  less  than  any  as- 
signed value,  however  small,  though  it  cannot  be  made 
absolutely  equal  to  zero. 

329.  The  ratio  between  two  incommensurable  quantities 
is  called  an  incommensurable  ratio. 

330.  As  the  treatment  of  Proportion  in  Algebra  depends 
upon  the  assumption  that  it  is  possible  to  find  fractions 
which  will  represent  the  ratios,  and  as  it  appears  that  no 
fraction  can  be  found  to  represent  the  exact  value  of  an 
incommensurable  ratio,  it  is  necessary  to  show  that  two 
incomn^ensurahle  ratios  are  equal  if  their  true  values  always 
lie  between  the  same  limits,  however  little  these  limits  differ 
from  each  other. 

Let  a :  h  and  c :  c?  be  two  incommensurable  ratios. 

Suppose  the  true  values  of  the  ratios  a  :  h  and  c  :  d  \\q  between 

!^  and  ^  .  Then  the  difference  between  the  true  values  of  these 
n  n  ,  ^ 

ratios  is  less  than  -,  however  small  the  value  of  -  may  be.  |  328. 

n  n 

But  since  -  can  be  made  to   approach   zero   at  pleasure,  -  can 
n  n 

be  made  less  than  any  assumed  difference  between  the  ratios. 

Therefore,  to  assume  any  difference  between  the  ratios  is  to  assume 

it  possible  to  find  a  quantity  that  for  the  same  value  of  —  shall  be 

1  ^^ 

both  greater  and  less  than  _  ;  which  is  a  manifest  absurdity. 
n 
Hence,  a-.h  =  c:d. 

331.  It  will  be  well  to  notice  that  the  word  limit  means 
a  fixed  value  from  which  another  and  variable  value  may 
be  made  to  differ  by  as  little  as  we  please ;  it  being  impos- 
sible, however,  for  the  difference  between  the  variable  value 
and  the  limit  to  become  absolutely  zero. 


RATIO.  281 


332.    A  ratio  will  not  he  altered  if  both  its  terms  he  multi- 
plied hy  the  same  numher. 

For  the  ratio  a.h  \a  represented  by  ^,  the  ratio  ttw,  :  mh  is  repre- 

,11      Tna         J    •        ma     a  x  j. 

sented  by  — -  ;  and  since  — —  =  -,  . .  ma  ■.mo=^a:o. 
mo  mo      0 


333.  A  ratio  will  he  altered  if  different  multipliers  of  its 
terms  he  taken ;  and  will  be  increased  or  diininished  accord- 
ing as  the  multiplier  of  the  antecedent  is  greater  or  less  than 
that  of  the  consequent. 

For,      ma :  nb  will  be  >  or  <  a :  6 

J.  ma  •    ^       ^  o,  f     na\ 

according  as  — -  is  >  or  <  -    =  — ~  ), 

nb  b\     no) 

as  ma  is  >  or  <  na, 

as  m    is  >  or  <  71. 

334.  A  ratio  of  greater  inequ/ility  will  be  diminished^  and 
a  ratio  of  less  inequality  increased  hy  adding  the  same  num- 
her to  both  its  terms. 

For,         a  +  x:b  +  X  is>  or  <  a-.b 
according  as  f        is  >  or  <  -, 

b  +  X  0 

as       ab  +  bx  is  >  or  <  ab  +  ax, 
as  tx  is  >  or  <  ax, 

as  J  is  >  or  <  a. 

335.  A  ratio  of  greater  incqualiiy  will  he  increased,  and  a 
ratio  of  less  inequality  diminished,  hy  subtracting  the  same 
numher  from  both  its  terms. 

For,  a  —  x:  b  —  x  will  be  >  or  <  a :  & 

J-  a  —  X  ■    ^       ^  a 

according  as is  >  or  <  -, 

b  —X  b 

as        ab  —  bx  is  >'or  <  ab  —  ax, 

as  ax  is  >  or  <  bx, 

as  a  is  >  or  <  6. 


282  ALGEBRA. 


336.  Katios  are  covipounded  by  taking  the  product  of  the 
fractions  that  represent  them. 

Thus,  the  ratio  compounded  of  o :  5  and  c-.d'vi  found  by  taking  the 

product  of  -  and  -  =  ^. 
0  d     hd 

The  ratio  compounded  of  a :  6  and  a :  6  is  the  duplicate  ratio  a^ :  h^, 

and  the  ratio  compounded  oi  a-.h,  a-.b,  and  a:b  is  the  triplicate  ratio 

a^:b\ 

337.  Eatios  are  compared  by  comparing  the  fractions  that 
represent  them. 

Thus,  a :  5  is  >  or  <  c :  f? 

according  as  -  is  >  or  <  — , 

b  d 

ad  ■    -^       ^  be 

as  —  IS  >  or  <  — , 

od  bd 

as  ad  is  >  or  <  be. 


Exercise  OXIII. 

1.  Write  down  the  ratio  compounded  of  3  :  5  and  8:7. 

Which  of  these  ratios   is   increased,   and  which  is 
diminished  by  the  composition? 

2.  Compound  the  duplicate  ratio  of  4  :  15  with  the  tripli- 

cate of  5 :  2. 

3.  Show  that  a  duplicate  ratio  is  greater  or  less  than  its 

simple  ratio  according  as  it  is  a  ratio  of  greater  or 
less  inequality. 

4.  Arrange  in  order  of  magnitude  the  ratios  3:4;  23  :  25 ; 

10:11;  and  15: 16. 

5.  Arrange  in  order  of  magnitude 

a-\-h:a  —  h  Sind  a^-\-h^ :  a^  —  h^,  U  a>h. 

Find  the  ratios  compounded  of: 

6.  3  :  5  ;  10  :  21 ;  14  :  15.  7.    7  :  9  ;  102  :  105  ;  15  :  17. 


RATIO.  283 


8.    — J ^^— Q  and ■ — . 

a^  —  a^x-\-  aar  —  of  a-\-x 

:r^  — G:r  a?  —  bx 

10.  a  +  5:a-5;  a^  _|.  ^,2 .  (^  _|_  ^^2 .  (^2  _  j2>^2 .  ^4  _  j4_ 

11.  Two  numbers  are  in  the  ratio  2  :  3,  and  if  9  be  added 

to  each,  they  are  in  the  ratio  3  :  4.     Find  the  num- 
bers. 
(Let  2x  and  3  a;  represent  the  numbers). 

12.  Show  that  the  ratio  a'.h  \&  the  duplicate  of  the  ratio 

a  -{-  c  '.  h  -\-  e,  ii  ^  =  ah. 

13.  Find  two  numbers  in  the  ratio  3  :  4,  of  which  the  sum 

is  to  the  sum  of  their  squares  in  the  ratio  of  7  to  50. 

14.  If  five  gold  coins  and  four  silver  ones  be  worth  as  much 

as  three  gold  coins  and  twelve  silver  ones,  find  the 
ratio  of  the  value  of  a  gold  coin  to  that  of  a  silver  one. 

15.  If  eight  gold  and  nine  silver  coins  be  worth  as  much 

as  six  gold  and  nineteen  silver  coins,  find  the  ratio 
of  the  value  of  a  silver  coin  to  that  of  a  gold  one. 

16.  There  are  two  roads  from  A  to  B,  one  of  them  14  miles 

longer  than  the  other ;  and  two  roads  from  B  to  C, 
one  of  them  8  miles  longer  than  the  other.  The  dis- 
tance from  A  to  B  is  to  the  distance  from  B  to  C,  by 
the  shorter  roads,  as  1  to  2 ;  by  the  longer  roads,  as 
2  to  3.     Find  the  distances. 

17.  "What  must  be  added  to  each  of  the  terms  of  the  ratio 

m  :  71,  that  it  may  become  equal  to  the  ratio p  '.  q'^ 

18.  A  rectangular  field  contains  5270  acres,  and  its  length 

is  to  its  breadth  in  the  ratio  of  31  :  17.  Find  its  di- 
mensions. 


284  ALGEBRA. 


Peopoetion. 

338.  An  equation  consisting  of  two  equal  ratios  is  called 
a  proportion ;  and  the  terms  of  the  ratios  are  called  propor- 
tionals. 

339.  The  algebraic  test  of  a- proportion  is  that  the  two 

fractions  which  represent  the  ratios  shall  be  equal. 

Thus,  the  ratio  a :  b  will  be  equal  to  the  ratio  c  :  cZ  if  -  =  -  ;  and 

b      d 
the  four  numbers  a,  b,  c,  d  are  called  proportionals,  or  are  said  to  be 

in  proportion. 

340.  If  the  ratios  a  :  h  and  c :  d  form  a  proportion,  the 
proportion  is  written  .  A  _    .  /J 

(read  the  ratio  of  a  to  5  is  equal  to  the  ratio  of  c  to  c?) 

or  a-.h  ::  c  :  d 

(read  a  is  to  h  in  the  same  ratio  as  c  is  to  d). 

The  first  and  last  terms,  a  and  d,  are  called  the  extremes. 

The  two  middle  terms,  h  and  <?,  are  called  the  means. 

341.  When  four  numhers  are  in  ^proportion,  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means. 

For,  if  a.b  ::  c:d, 

then  -  =  -. 

b      d 

By  multiplying  by  bd,   ad  =  be. 

342.  If  the  product  of  two  numhers  he  equal  to  the  product 
of  two  others,  either  two  m^ay  he  m^ade  the  extremes  of  a  pro- 
portion and  the  other  two  the  means. 

For,  if  ad  =  be, 

by  dividing  by  6(/,  ^=^, 

od      bd 

.'.  a:  b  ::  c:  d. 


PROPORTION.  285 


343.  The  equation  ad  =  be  gives 

be       ,       ad 

a  c 

so  that  an  extreme  may  be  found  by  dividing  the  product 
of  the  means  by  the  other  extreme ;  and  a  mean  may  be 
found  by  dividing  the  product  of  the  extremes  by  the  other 
mean. 

344.  If  four  quantities,  a,  6,  c,  d,  be  in  proportion,  they 
will  be  in  proportion  by  : 

I.   Inversion. 

That  is,  b  will  be  to  a  as  c?  is  to  c. 

For,  if        a:h  ::  c:d. 


then 

a 

b- 

c 

^5' 

and 

-r 

-'-r 

or 

h 
a 
.:b:a 

_d 
c' 
:.d:c. 

345. 

II. 

Composition. 

That 

is, 

a-{-b  will  be  to  5  as  c  +  c?  is 

toe?. 

For,  if 

a:b:: 

c:d, 

then 

a 

c 

and 

a      ^      c       ^ 

i  +  l-3  +  l' 

or 

a+b     c+d 
b     ~     d    ' 

.'.a  +  b:b  ::  C  +  d:d. 

346.   III.   Division. 

That  is,     a  —  b  will  be  to  5  as  c  —  c?  is  to  d. 

For, 

if        a.b  ::  c:d, 

then 

a      c 
b-^d' 

286  ALGEBRA. 


and  ■. —  1  =  --  —  1, 

h  d 

a—hc—d 
.'.  a  —  b  :  h  : :  c  —  d:  d. 

347.  IV.   Composition  and  Division. 

That  is,  a-{-b  will  he  to  a  —  b  SiS  c-{-d  is  to  c  —  d. 

For,  from  II.,        £l±A==^±1^ 
h  d    ' 

and  from  III..  9lz:±=<LizA, 

h  d 

By  dividing,  a  +  h_c±d 

^  ^'  a-h      c-d 

.•.a  +  h:  a—b  ::  c  +  d:  c  —  d. 

348.  When  the  four  quantities  a,  b,  c,  d  are  all  of  the 
same  kind,  they  will  be  in  proportion  by  : 

V.   Alternation. 

That  is,  a  will  be  to  c  as  5  is  to  d. 

For,  if  a:b  .:  c:d, 


then 

a      c 
b~d' 

By  multiplying  by  -, 
c 

ah      be 
be  ~  cd' 

or 

a      b 
c  ~  d' 

.'.a:  c  ::  h:d. 

349.    From  the  proportion  a\c\\  b-.d  may  be  obtained 
by: 

VI.  Composition.    a-\-c\c\\b-\-d\d. 

VII.  Division.  a  —  c  :  c  : :  b  —  d :  d. 

VIII.  Composition  and  Division.  a'\-c:a~-€  ::b-\-d:b  —  d. 


PROPORTION.  287 


350.  In  a  series  of  equal  ratios,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

For,  if  ^-  =  ^.  =  ^  =  £, 

b      d     f     h 

r  may  be  put  for  each  of  these  ratios. 

Then    -^-  =  r,  l=r,  1  =  r,  5^.  =  r, 
b  d  f  h 

.'.  a  =  br,  c  =  dr,  e  =fr,  g  =  hr. 

.-.  a  +  c  +  e  +  g  ==  {b  +  d  +f  +  h)  r. 

, a+c+e+g_    _a 

b  +  d+f+h  ~    ~b' 

.'.a  +  c  +  e  +  g  :  b  +  d  +/+  h  ::  a  :  b. 

In  like  manner,  it  may  be  ehown  that 

ma  +  nc  +pe  +  qg  :  mb  +  nd  +  pf  +  qh  : :  a  :  b. 

351.  If  a,  h,  c,  d  be  in  continued  proportion,  that  is,  if 
a:b  =  b  :c=c:d,  then  will  a:  c  —  a^:b^  and  a  :  d  =  a^  :  b^. 


For, 

a      b      c 
b~c~d' 

Hence, 

a      b      a      a 

b^c~b^F 

a     a« 

.•.a:c  =  a«:6«. 

So 

a 
b 

V*  v^  -^  V*  v^ 
"^c^d-b^b^b 

a     a? 

d-w 

:.a.d  =  a?.b\ 

352.  If  a,  b,  c  be  proportionals,  so  that  a:b  ::b:  c,  then 
b  is  called  a  mean  proportional  between  a  and  c,  and  c  is 
called  a  third  proportional  to  a  and  b. 


288 


ALGEBRA. 


lia:h 


b  :  c,  then  b  = 

-y/ac. 

For,  if 

a:  b  : :  b  :  c. 

then 

a      b 
b      c' 

and 

b^=aG, 

.•.  b  =  y/ac. 

353.    The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 


For,  if 

a:b: 

■.c:d. 

and 

h:l: 

:  m:n, 

then 

ace 
b~d'   J 

g     k      m 

Hence,  by  finding  the  product  of  the  left  members,  and  also  of  the 
ight  members  of  these  equations, 

aek 

bfi- 

cgm 
dhn 

■.aek-.bfl: 

cgm :  dhn. 

354.    lyike  powers,  or  like  roots,  of  the  tei^ms  of  a  propor- 
tion are  in  proportion. 

For,  if  a-.b  ::  c:d, 

then  J  =  l 


By  raising  both  sides  to  the  nth  power, 
o^  _  c^ 

By  extracting  the  nth  root, 
a^      c^ 


b^     d^ 


.'.a^:b^ 


A.  A 


PROPORTION.  289 


355.  If  two  quantities  he  increased  or  diminished  hy  like 
parts  of  each,  the  results  will  he  in  the  same  ratio  as  the 
quantities  themselves. 

a  _  \        n) 


For. 


b       /  .  .  ,n  ^  ^ 


that  is. 


n 


356.  The  laws  that  have  been  established  for  ratios 
should  be  rememembered  when  ratios  are  expressed  in 
their  fractional  form. 


x—1      x^-^x  —  2' 


(1)  Solve: 

By  g  347 

and  this  equation  is  satisfied,  when  x  =  0] 

A-   -A-       V     2a;*  1  1 

or,  dividincr  by  — ,  = , 

°    "^     2'  a;  +  l      2-x 

(2)  J^  a  :  h  : :  c  :  d,  show  that 

a^-i-ab:P-ah::(^  +  cd:d:'-cd. 


If 

a      c 
b~d' 

then 

a  +  b     c  +  d 
a  —  b      c  —  d' 

and 

a         c 
~b     -d' 

.     a    ^^a  +  b       c        c  +  d, 
■•-b'^a-b      ~d^^c-d' 

that  is, 

a^  +  ab      (?  +  cd 
b^-ab      d^-cd' 

or 

a^  +  ab:b^-ab::(^  +  cd:d^-cd. 

^34: 


p53. 


290  ALGEBRA. 


(3)    When  a  \  h  \ :  c  :  d,  and  a  is  the  greatest  term,  show  that 
a-\-  di&  greater  than  h  -{-  c. 

Since  ^  =  -.  ^nd  a  >  5, 

.-.  5  >  d 
Also,  since  a-h^Cj-d^  ^  3^g 

and  h>  d, 

: .  a  —  h  >  c  —  d. 

By  adding,  h  +  d  =1)  +  d, 

a  +  d>h  -\-  c. 


Exercise  CXIV. 
li  a:h  '.'.  c'.d,  prove  that : 

1.  ma  -.nh::  mc  :  nd.  4.    a^  -.I/wc^:  d^. 

2.  ^a-{-h  -.h  ::  Zc-\-d:  d.  5.    a:  a-{-h  :  :  c  :  c-\-d. 

3.  a-\-2h  \h  ::  c-\-2>d\d.  6.    a  :  a  —  h  :  \  c  :  c  —  d. 

7.  wz(2  +  72^  :  Tna  —  nh  -.:  mc  +  nd :  mc  —  nd. 

8.  2a4-3Z>:3a-45  ::  2c  +  3c^:3c-4d 

9.  ??2a^  +  nc^ :  mh^  +  nd^  \\  c?  :  h^. 

10.    ?^a^  +  nah  -\-p¥  :  mc^  +  '^^'^  +^<i^  :  :  Z*^ :  c?^. 

If  a  :  h  :  :  h  :  c,  prove  that : 
11.    a-i-h  •.b-{-c  :  :  a:h.  12.    c^ ^ah  \J? -\-'bc  w  a\e. 

13.    a:Z»  ::  (a  +  ^f  :  (^  +  ^)'. 

14.  When  a,  Z>,  and  c  are  proportionals,  and  a  the  greatest, 

show  that  a+  c  >  25. 

15.  If^~"y  =  y  "'^^g^Z^^  and  x,  y,  z  be  unequal,  then 

I  m  n 


PROPORTION.  291 


16.  Find  X  when  x-^b  :2x  —  2>  :\  bx-\-l  :^x  —  ?>. 

17.  Find  X  when  x-{-a:2x  —  h  :  :  2tx-\-b  :^x  —  a. 

18.  Find  X  when  -^/x  +  V6  :  V.t  —  V6  ::  a:h. 

19.  Find  a;  and  y  when  a; :  27  :  :  y  :  9,  and  a; :  27  :  :  2  :  a:— y. 

20.  Find  a;  and  y  when  a;4-y+l:^  +  y  +  2::  6:7,  and 

when2/4-2:r:?/-2a:::  12a;4-6y-3  :  6y-12a:-l. 

21.  Find  X  when  a;2_4^-j-2  :  a:2_2a;-l  : :  x^-^x  :  a:2_2a;-2. 

22.  A  railway  passenger  observes  that  a  train  passes  him, 

moving  in  the  opposite  direction,  in  2  seconds ;  but 
moving  in  the  same  direction  with  him,  it  passes  him 
in  30  seconds.     Compare  the  rates  of  the  two  trains. 

23.  A  and  B  trade  with  different  sums.     A  gains  $200  and 

B  loses  $50,  and  now  A's  stock  :  B's  :  :  2  :  J.  But,  if 
A  had  gained  $100  and  B  lost  $85,  their  stocks 
would  have  been  as  15  :  3  J.  Find  the  original  stock 
of  each. 

24.  A  quantity  of  milk  is  increased  by  watering  in  the  ra- 

tio 4  : 5,  and  then  3  gallons  are  sold  ;  the  remainder 
is  mixed  with  3  quarts  of  water,  and  is  increased  in 
the  ratio  6  :  7.  How  many  gallons  of  milk  were  there 
at  first? 

25.  In  a  mile  race  between  a  bicycle  and  a  tricycle  their 

rates  were  as  5  :  4.  The  tricycle  had  half  a  minute 
start,  but  was  beaten  by  176  yards.  Find  the  rates 
of  each. 

26.  The  time  which  an  express-train  takes  to  travel  180 

miles  is  to  that  taken  by  an  ordinary  train  as  9  :  14. 
The  ordinary  train  loses  as  much  time  from  stopping 
as  it  would  take  to  travel  30  mjles ;  the  express-train 
loses  only  half  as  much  time  as  the  other  by  stopping, 
and  travels  15  miles  an  hour  faster.  What  are  their 
respective  rates? 


292  ALGEBRA. 


27.  A  line  is  divided  into  two  parts  in  the  ratio  2:3,  and 

into  two  parts  in  the  jratio  3:4;  the  distance  be- 
tween the  points  of  section  is  2.  Find  the  length  of 
the  line. 

28.  A  railway  consists  of  two    sections ;    the  annual    ex- 

penditure on  one  is  increased  this  year  5%,  and  on 
the  other  4%,  producing  on  the  whole  an  increase  of 
^TU%-  Compare  the  amount  expended  on  the  two 
sections  last  year,  and  also  this  year. 

29.  When  a,  b,  c,  d  are  proportional  and  unequal,  show  that 

no  number  x  can  be  found  such  that  a-\-x,  b-j-x, 
c-}-x,  d-\-x  shall  be  proportionals. 

Variation. 

357.  Two  quantities  may  be  so  related  that,  when  one  has 
its  value  changed,  the  other  will,  in  consequence,  have  its 
value  changed. 

Thus,  the  distance  travelled  in  a  certain  time  will  be  doubled  if 
the  rate  be  doubled.  The  time  required  for  doing  a  certain  quantity 
of  work  will  be  doubled  if  only  half  the  number  of  workmen  be 
employed. 

358.  Whenever  it  becomes  necessary  to  express  the  gen- 
eral relations  of  certain  kinds  of  quantities  to  each  other, 
without  confining  the  inquiry  to  any  2^ci'riicular  values  of 
these  quantities,  it  will  usually  be  sufficient  to  mention  two 
of  the  terms  of  a  proportion.  In  all  such  cases,  however, 
four  terms  are  always  implied. 

Thus,  if  it  be  said  that  the  weight  of  water  is  proportional  to  its 
volume,  or  varies  as  its  volume,  the  meaning  is,  that  one  gallon  of 
water  is  to  any  number  of  gallons  as  the  weight  of  one  gallon  is  to  the 
weight  of  the  given  number  of  gallons. 


VARIATION.  293 


359.  Quantities  used  in  a  general  sense,  as  distance,  time, 
weight,  volume,  to  which  particular  values  may  be  assigned, 
are  denoted  by  capital  letters.  A,  B,  C,  etc. ;  while  as- 
signed values  of  thdse  quantities  may  be  denoted  by  small 
letters,  a,  h,  c,  etc.  The  letters  A,  B,  Cwill  be  understood 
to  represent  any  numerical  values  that  may  be  assigned  to 
the  quantities ;  and  when  two  such  letters  occur  in  an  ex- 
pression they  will  be  understood  to  represent  any  corre- 
sponding numerical  values  that  may  be  assigned  to  the  two 
quantities. 

360.  "When  two  quantities  A  and  B  are  so  connected 
that  their  ratio  is  constant,  that  is,  remains  the  same  for  all 
corresjponding  values  of  A  and  B,  the  one  is  said  to  vary  as 
the  other ;  and  this  relation  is  expressed  by  ^  oc  i>  (read 
A  varies  as  B). 

Thus,  the  area  of  a  triangle  with  a  given  base  varies  as  its  altitude ; 
for,  if  the  altitude  be  changed,  the  area  will  be  changed  in  the  same 
ratio. 

A 

If  this  constant  ratio  be  denoted  by  w,  then  -—  —  m,  or  A 

^mB.  ^ 

From  this  equation  ni  may  be  found  when  two  corre- 
sponding values  of  A  and  B  are  known. 

361.  When  two  quantities  are  so  connected  that  if  one 
be  changed  in  any  ratio,  the  other  will  be  changed  in  the 
inverse  ratio,  the  one  is  said  to  vary  inversely  as  the  other. 

Thus,  the  time  required  to  do  a  certain  amount  of  work  varies  in- 
versely as  the  number  of  workmen  employed  ;  for,  if  the  number  of 
workmen  be  doubled,  halved,  or  changed  in  any  ratio,  the  time  re- 
quired will  be  halved,  doubled,  or  changed  in  the  inverse  ratio. 

362.  If  A  vary  inversely  as  B,  two  values  of  A  have  to 
each  other  the  inverse  ratio  of  the  two  corresponding  values 


294  ALGEBRA. 


Hence,  the  product  A£  is  constant,  and  may  be  denoted 
by  on.     That  is,  A£  =  m. 

If  any  two  corresponding  values  of  A  and  JB  be  known, 
the  constant  m  may  be  found. 

The  equation  AB  =  m  may  be  written  A=-—,  and  as  ?m 
is  constant,  A  is  said  to  vary  as  the  o^eciprocal  of  JB,  or 

363.  The  two  equations, 

A  =  mJB  (for  direct  variation), 
A  =  ---  (for  inverse  variation), 

furnish  the  simplest  method  of  treating  Variation. 
If  A^^viBC,  A  is  said  to  ysltj  joinify  as  B  and  C. 

If  ^  =  — ^,  A  is  said  to  vary  directly  as  B  and  inversely 

364.  The  following  results  are  to  be  observed : 

I.  If  ^  oc  ^  and  i?  oc  C,  then  A  o:  C. 

For  A  =  7nB,  where  m  is  constant, 

and  B  =  nC,    where  n  is  constant. 

.'.  A  =  mnC. 
.'.  A  cc  C,  since  mn  is  constant. 

In  like  manner,  if  ^  cc  ^  and  B  cc  -,  then  A  cc  —. 

C  C 

II.  If  ^  oc  C'and  Bo:C,  then  ^  ±  i?  oc  C'.  and  VXg  oc  C. 

For  A=^  7nC,  where  m  is  constant, 

and  B  =■  nC,  where  n  is  constant. 

.•.A±B^{m±n)C. 
.-.  A±  B  cc  C  since  m  ±  w  is  constant. 


Also,     VAB  =  VmCx  nO=  y/mnC^^  C^fmn. 
: .  ^ AB  oc  (7  since  y/mn  is  constant. 


VARIATION.  295 


III.   U  A  a:  B  siiid  CcxiD,  then  AC  ccJBD. 

For  A  =  mB,  where  m  is  constant, 

(7=  nD,    where  n  is  constant. 
.-.  AC=  mnBD. 
.'.  AC (K  BD,  since  mn  is  constant. 


IV.    liAazB  then  J"  oc  ^". 

For  A  =  mB,  where  m  is  constant. 

.'.  A^  oc  ^"     since  m"  is  constant. 


V.  If  J.  cc  ^  when  C  is  unchanged,  and  A  a:  C  when  ^ 
is  unchanged,  then  A  cc  ^Cwhen  both  B  and  C  change. 

For        J.  =  mB,  when  -S  varies  and  C  is  constant. 

Here,  m  is  constant  and  cannot  contain  the  variable  B 

.'.  A  must  contain  B,  but  no  other  power  of  B. 

Again,     A  =  nC,  when  C  varies  and  B  is  constant. 

Here,  n  is  constant  and  cannot  contain  the  variable  C, 

.'.  A  must  contain  C,  but  no  other  power  of  C. 

Hence,  A  contains  both  B  and  C,  but  no  other  powers  of  B 

and  C,  and  therefore, 

A 
-— -  =j3,  or  A  =pBC,  where  p  is  constant. 

.'.  A  cc  BCy    since  p  is  constant. 

In  like  manner,  it  may  be  shown  that  if  A  vary  as  each 
of  any  number  of  quantities  B,  C,  D,  etc.,  when  the  rest 
are  unchanged,  then  when  they  all  change,  A  oc  BCD,  etc. 

Thus,  the  area  of  a  rectangle  varies  as  the  base  when  the  altitude 
is  constant,  and  as  the  altitude  when  the  base  is  constant,  but  as  the 
product  of  the  base  and  altitude  when  both  vary. 

The  volume  of  a  rectangular  solid  varies  as  the  length  when  the 
width  and  thickness  remain  constant;  as  the  width  when  the  length 
and  thickness  remain  constant ;  as  the  thickness  when  the  length  and 
width  remain  constant ;  but  as  the  product  of  length,  breadth,  and 
thickness  when  all  three  vary. 


296  ALGEBRA. 


(1)  If  A  vary  inversely  as  B,  and  when  A  =  2  the  corre- 
sponding value  of  £  is  36,  find  the  corresponding 
value  of  £  when  ^  =  9. 


or 

m  = 

B' 
=  AB, 

=  2  X  36  = 

72. 

And  if  9  and  ' 

72  be  substituted  for  A  and 

m  respectively 

in 

the  result  is 

A  = 
9  = 

m 

~  B' 

72 
=  — . 

B 

•' 

.'.B  = 

=  72. 
=  8.  Ans. 

(2)  The  weight  of  a  sphere  of  given  material  varies  as  its 
volume,  and  its  volume  varies  as  the  cube  of  its  diam- 
eter. If  a  sphere  4  inches  in  diameter  weigh  20 
pounds,  find  the  weight  of  a  sphere  5  inches  in  diam- 
eter. 

Let     TT  represent  the  weight, 
V  represent  the  volume, 
D  represent  the  diameter. 
Then  TTocFand  Voz  D^, 

.-.  TFoc  D^. 
Put     W=  mD^, 
then,  since  20  and  4  are  corresponding  volumes  of  W  and  Z), 
20  =  m  X  64, 

.-.  when  Z>  =  5,  Tr=  y^  of  125  =  39 J^. 

Exercise  CXV. 

1.  If^oc.5,  and  ^  =  4when.5  =  5,  find  ^  when  ^  =  12. 

2.  If  ^  oc  B,  and  when  B  =  ^,  A  =  ^,  find  A  when  B^l. 

3.  If  A  vary  jointly  as  B  and  C,  and  3,  4,  5  be  simulta- 

neous values  of  A,  B,  C,  find  A  when  ^  =  0=  10. 


VARIATION.  297 


^4.    If  ^  oc  -^,  and  when  A  =  10,  B  =  2,  find  the  value  of 
£  when  A  =  4:. 

5.  If  ^  oc  ~  and  when  A  =6,  B  =  4:,  and   (7=3,  find 

the  value  of  A  when  JB  =  5  and  C=7. 

6.  If  the  square  of  X  vary  as  the  cube  of  F,  and  X=  3 

when  F=  4,  find  the  equation  between  Xand  Y. 

7.  If  the  square  of  X  vary  inversely  as  the  cube  of  Y,  and 

X=2  when  F=  3,  find  the  equation  between  X 
and  Z 

8.  If  -^  vary  as  X  directly  and  Y  inversely,  and  if  when 

Z=2,  X=3,  and  F=4,  find  the  value  of  ^  when 
X=15andF-:8. 

9.  li  A  cc  B-{-c  where  c  is  constant,  and  if  ^  =  2  when  B 

=  1,  and  if  ^  =  5  when  B  =  2,  find  A  when  ^  =  3. 

10.  The  velocity  acquired  by  a  stone  falling  from  rest  varies 

as  the  time  of  falling;  and  the  distance  fallen  varies 
as  the  square  of  the  time.  If  it  be  found  that  in  3 
seconds  a  stone  has  fallen  145  feet,  and  acquired  a 
velocity  of  96f  feet  per  second,  find  the  velocity  and 
distance  at  the  end  of  5  seconds. 

11.  If  a  heavier  weight  draw  up  a  lighter  one  by  means  of 

a  string  passing  over  a  fixed  wheel,  the  space  de- 
scribed in  a  given  time  will  vary  directly  as  the 
difference  between  the  weights,  and  inversely  as  their 
sum.  If  9  ounces  draw  7  ounces  through  8  feet  in  2 
seconds,  how  high  will  12  ounces  draw  9  ounces  in 
the  same  time  ? 

12.  The  space  will  vary  also  as  the  square  of  the  time. 

Find  the  space  in  Example  11,  if  the  time  in  the  lat- 
ter case  be  3  seconds. 


298  ALGEBRA. 


13.  Equal  volumes  of  iron  and  copper  are  found  to  weigh 

77  and  89  ounces  respectively.  Find  the  weight  of 
10  J  feet  of  round  copper  rod  when  9  inches  of  iron 
rod  of  the  same  diameter  weigh  SIj^q-  ounces. 

14.  The  square  of  the  time  of  a  planet's  revolution  varies  as 

the  cube  of  its  distance  from  the  sun.  The  distances 
of  the  Earth  and  Mercury  from  the  sun  being  91  and 
85  millions  of  miles,  find  in  years  the  time  of  Mer- 
cury's revolution. 

15.  A  spherical  iron  shell  1  foot  in  diameter  weighs  -^^g-  of 

what  it  would  weigh  if  solid.  Find  the  thickness 
of  the  metal,  it  being  known  that  the  volume  of  a 
sphere  varies  as  the  cube  of  its  diameter. 

16.  The  volume  of  a  sphere  varies  as  the  cube  of  its  diame- 

ter. Compare  the  volume  of  a  sphere  6  inches  in 
diameter  with  the  sum  of  the  volumes  of  three  spheres 
whose  diameters  are  3,  4,  5  inches  respectively. 

17.  Two  circular  gold  plates,  each  an  inch  thick,  the  diam- 

eters of  which  are  6  inches  and  8  inches  respectively, 
are  melted  and  formed  into  a  single  circular  plate 
1  inch  thick.  Find  its  diameter,  having  given  that 
the  area  of  a  circle  varies  as  the  square  of  its  diameter. 

18.  The  volume  of  a  pyramid  varies  jointly  as  the  area  of 

its  base  and  its  altitude.  A  pyramid,  the  base  of  which 
is  9  feet  square,  and  the  height  of  which  is  10  feet,  is 
found  to  contain  10  cubic  yards.  What  must  be  the 
height  of  a  pyramid  upon  a  base  3  feet  square,  in 
order  that  it  may  contain  2  cubic  yards  ? 


CHAPTER    XXI. 
Series. 

365.  A  succession  of  numbers  whicli  increase  or  decrease 
according  to  some  fixed  law  is  called  a  series  |  and  the  suc- 
cessive numbers  are  called  the  terms  of  the  series. 

Thus,  by  executing  the  indicated  division  of ,  the  series  1  +x 

1  —  x 

+  a^  +  a^  + is  obtained,  a  series  that  has  an  unlimited  number  of 

terms. 

366.  A  series  that  is  continued  indefinitely  is  called  an 
infinite  series ;  and  a  series  that  comes  to  an  end  at  some 
particular  term  is  called  a  finite  series. 

367.  When  a:  is  <  1,  the  more  terms  we  take  of  the  infi- 
nite series  1 -\- x -\- 3^ -{■  3? -{■ ,  obtained  by  dividing  1  by 

1  —  x,  the  more  nearly  does  their  sum  approach  to  the  value 

of     1 
1—x 

1  13 

Thus,  if  X  =  J,  then = ■  =  -,  and  the  series  becomes  1  +  i- 

1-a;      1-J      2 

+  i  +  ^T+ '  ^  ^^^  which  cannot  become  equal  to  f  however  great 

the  number  of  terms  taken,  but  which  may  be  made  to  differ  from  f 
by  as  little  as  we  please  by  increasing  indefinitely  the  number  of 
terms. 

368.  But  when  x  is  >  1,  the  more  terms  we  take  of  the 

series  l+5;-f-^  +  ^+ the  more  does  the  sum  of  the 

series  diverge  from  the  value  of . 

X       X 

Thus,  if  cc  =  3,  then  :; =  ;; =  — ,  and  the  series  becomes  1  + 

1-x     1-3         2 
3  +  9  +  27  + ,  a  sum  which   diverges   more   and  more   from   —  J, 


300  ALGEBRA. 


the  more  terms  we  take,  and  which  may  be  made  to  increase  indefi- 
nitely by  increasing  indefinitely  the  number  of  terms  taken. 

369.  A  series  wliose  sum  as  the  number  of  its  terms  is  in- 
definitely increased  approaches  some  fixed  finite  value  as  a 
limit  is  called  a  converging  series ;  and  a  series  whose  sum 
increases  indefinitely  as  the  number  of  its  terms  is  increased, 
is  called  a  diverging  series. 

370.  When  x  =  l,  the  division  of  1  hj  1  —  x,  that  is,  of 
1  by  0,  has  no  meaning,  according  to  the  definition  of  divi- 
sion; and  any  attempt  to  divide  by  a  divisor  that  is  equal 
to  zero  leads  to  absurd  results. 

Thus,  8  +  4  =  8+4; 

by  transposing,  8  —  8  =  4  —  4; 

or,  dividing  by  4  —  4,        2  =  1;         a  manifest  absurdity. 

371.  When  x=l  very  nearly,  then  the  value  of 


will  be  very  great,  and  the  sum  of  the  series  1  +  ^  +  ^  + 

01? -\- will  become  greater  and  greater  the  more  terms  we 

take.     Hence,  by  making  the  denominator  1—x  approach 
indefinitely  to  zero,  the  value  of  the  fraction  - — —  may  be 

J.        X 

made  to  increase  at  pleasure. 

372.  If  the  symbol  o  be  used  to  denote  a  quantity  that 
is  less  than  any  assignable  quantity,  and  that  may  be  con- 
sidered to  decrease  without  limit,  not,  however,  becoming 
0,  and  the  symbol  oo  be  used  to  denote  a  quantity  that  is 
greater  than  any  assignable  quantity,  and  that  may  be  con- 
sidered to  increase  without  limit,  then 

O 

In  the  same  sense  ^=oo,  where  a  represents  any  value 
that  may  be  assigned. 


SERIES.  301 


1—3^ 

373.  If  X  in  the  fraction  :; —  be  equal  to  1,  the  numer- 

1  —X 

ator  and  denominator  will  each  become  0,  and  the  fraction 
will  assume  the  form  -. ' 

374.  If,  however,  x  in  this  fraction  approach  to  1  as  its 
limit,  then  the  denominator  1  —  x,  inasmuch  as  it  has  some 
value,  even  though  less  than  any  assignable  value,  may  be 
used  as  a  divisor,  and  the  result  is  l-\-x  +  x^-{-a^-\-x*. 
Hence,  it  is  evident  that  though  both  terms  of  the  fraction 
become  smaller  and  smaller  as  1  —  ^  approaches  to  0,  still 
the  numerator  becomes  more  and  more  nearly  five  times  the 
denominator. 

It  may  be  remarked  that  when  the  symbol  ^  is  obtained  for  the 
value  of  the  unknown  quantity  in  a  problem,  the  meaning  is  that  the 
problem  has  no  definite  solution,  but  that  its  conditions  are  satisfied 
if  any  value  whatever  be  taken  for  the  required  quantity  ;  and  if  the 
symbol  %,  in  which  a  denotes  any  assigned  value,  be  obtained  for 
the  value  of  the  unknown  quantity,  the  meaning  is  that  the  condi- 
tions of  the  problem  are  impossible. 

375.  The  number  of  different  series  is  unlimited,  but  the 
only  kinds  of  series  that  can  be  considered  in  a  work  of  this 
character  are  Arithmetical,  Geometrical,  and  Harmonical 
Series. 

Arithmetical  Series. 

376.  A  series  in  which  the  difference  between  any  two 
adjacent  terms  is  equal  to  the  difference  between  any  other 
two  adjacent  terms,  is  called  an  Arithmetical  Series  or  an 
Arithmetical  Progression. 

377.  The  general  representative  of  such  a  series  will  be 

a,  a-{-d,  a -{-2d,  a-{-^d , 

in  which  a  is  the  first  term  and  d  the  common  difference ; 


302  ALGEBRA. 


and  the  series  will  be  increasing  or  decreasing  according  as 
d  is  positive  or  negative. 

378.  Since  each  succeeding  term  of  the  series  is  obtained 
by  adding  d  to  the  preceding  term,  the  coefficient  of  d  will 
always  be  1  less  than  the  number  of  the  term,  so  that  the 

nth  term  ^=  a -\-  (n  —  V)  d. 

If  the  nth  term  be  denoted  by  I,  this  equation  becomes 

l=a  +  {n-l)d.  (1) 

379.  The  arithmetical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
an  arithmetical  series. 

380.  If  a  and  h  denote  two  numbers,  and  A  their  arith- 
metical mean,  then,  by  the   definition  of  an  arithmetical 

.■.A  =  ^.  (2) 

381.  Sometimes  it  is  required  to  insert  several  arithmeti- 
cal means  between  two  numbers. 

If  772,  z=  the  number  of  means,  then  w  +  2  =  n,  the  whole 
number  of  terms ;  and  if  m  +  2  be  substituted  for  n  in  the 

^^^^*i°^  '      l=a-{-(n-l)d, 

the  result  is  I  =  a -\- {m -\- 1)  d. 

By  transposing  a,     l—a  =  {m-\-\)d, 

Thus,  if  it  be  required  to  insert  six  means  between  3  and  17,  the 

17  —  3 
value  of  d  is  found  to  be =  2  ;  and  the  series  will  be  3,  5,  7,  9, 

6  +  1 
11,  13,  15,  17. 


SERIES.  303 

382.  If  I  denote  the  last  term,  a  the  first  term,  n  the 
number  of  terms,  d  the  common  diiference,  and  s  the  sum 
of  the  terms,  it  is  evident  that 

s=      a    +(a+^)  +  (a+2c?)  + +  (/-c?)+      Z,  or 

_s  -=      I     -\-ll-d)-{-{l-2d)  + +  («  +  c?)+      a 

.•.2s-(a+0  +  («+0  +  («+0     + +  («  +  0  +(«+0 

=  72  (a  +  0 

383.  From  the  two  equations, 

l  =  a-\-{n-l)d,  (1) 

any  one  of  the  quantities  a,  d,  I,  n,  s  may  be  found  when 
three  are  given. 

Ex.  Find  n  when  d,  I,  s  are  given. 

From  (1),  a=l-{n-l)d. 

2s -In 


From  (2), 

Therefore,      l  —  {n  —  l)d 


n 
2s -In 


n 
In  —  dn^  +  dn  =  2s  —  ln. 


.'.d7v'-{2l  +  d)n==-2s, 
,'Adhi'  -{)  +  {2l  +  df  =  {2l  +  df-Sds, 


.■.2dn-{2l  +  d)  =  ±  V(2 l  +  df-8 ds, 

•  ^_2l  +  d±Vi2l  +  dY-8ds 
2d 

Note.  The  table  on  the  following  page  contains  the  results  of  the 
general  solution  of  all  possible  problems  in  arithmetical  series.  The 
student  is  advised  to  work  these  out,  both  for  the  results  obtained 
and  for  the  practice  gained  in  solving  literal  equations  in  which  the 
unknown  quantities  are  represented  by  other  letters  than  x,  y,  z. 


304 

No. 


ALGEBRA. 


Given. 


Required. 


Results. 


a  d  n 
ads 
a  n  s 
d  n  s 


=  a  +  (n  —  1)  d 


id±^/[2ds  +  {a-^df]. 


{n-l)d 


2 


a  d  I 
a  n  I 
d  n  I 


s=  ln[2a  +  {n  —  l)d]. 

*         2  2d    ' 

s=il  +  a)^. 

s=hn[2l-{n-l)d]. 


10 
11 
12 


d  n  I 
d  n  s 
d  I  s 
n  I  s 


I -ill-  1)  d. 

1  _  (^  —  1)  c? 
n  2 


id±  V{l  +  idf-2ds. 


13 
14 
15 
16 

17 
18 
19 
20 


a  n  I 

a  n  s 

a  I  s 

n  I  8 


a  d  I 
ads 
a  I  s 
d  I  s 


1  —  a 
?i-l' 

2  (g  -  an) 
n  {n  —  1) 

2s -I -a 
2{nl-s) 
n{n  —  l) 


I  —  a 


+  1. 


d-2a±V{2a-df  rSds 
2d 
2s 
I  +  a 

2l  +  d±  V{2l  +  dY-8d^ 
2d 


SERIES.  305 


Exercise  CXVI. 

1.  Find  the  thirteenth  term  of  5,  9,  13 

ninth  term  of  —  3,  —  1,  1 

tenth  term  of  —  2,  —  5,  —  8 

eighth  term  of  a,  a-{-Sb,  a-^-Q h 

fifteenth  term  of  1,  f,  -f- ~\ 

thirteenth  term  of  —  48,  —44,-40 J 

2.  The  first  term  of  an  arithmetical  series  is  3,  the  thir- 

teenth term  is  55.     Find  the  common  difference. 

3.  Find  the  arithmetical  mean  between :    (a.)  3  and  12 ; 

(h.)  -  5  and  17 ;  {c.)  o?^ah~  b^  and  0^-06  +  b\ 

4.  Insert  three  arithmetical  means  between  1  and  19 ;  and 

four  means  between  —  4  and  17. 

5.  The  first  term  of  a  series  is  2,  and  the  common  differ- 

ence J.     What  term  will  be  10  ? 

6.  The  seventh  term  of  a  series,  whose  common  difference 

is  3,  is  11.     Find  the  first  term. 

7.  Find  the  sum  of 

5  +  8  +  11-1- to  ten  terms. 

—  4  —  1  +  2  + to  seven  terms. 

a  +  4a  +  7a  + to  n  terms. 

-|  +  -j^  +  ^+ to  twenty-one  terms. 

1  +  21  +  4-J  + to  twenty  terms. 

8.  The  sum  of  six  numbers  of  an  arithmetical  series  is  27, 

and  the  first  term  is  1.     Determine  the  series. 

9.  How  many  terms  of  the  series  —5  —  2  +  1+ must 

be  taken  so  that  their  sum  may  be  63. 

10.    The  first  term  is  12,  and  the  sum  of  ten  terms  is  10. 
Find  the  last  term. 


306  ALGEBRA. 


11.  The  arithmetical  mean  between  two  numbers  is  10,  and 

the  mean  between  the  double  of  the  first  and  the 
triple  of  the  second  is  27.     Find  the  numbers. 

12.  Find  the  middle  term  of  eleven  terms  whose  sum  is  66. 

13.  The  first  term  of  an  arithmetical  series  is  2,  the  common 

diiference  is  7,  and  the  last  term  79.  Find  the  num- 
ber of  terms. 

14.  The  sum  of  fifteen  terms  of  an  arithmetical  series  is  600, 

and  the  common  difierence  is  5.     Find  the  first  term. 

15.  Insert  ten  arithmetical  means  between  —  7  and  114. 

16.  The  sum  of  three  numbers  in  arithmetical  progression 

is  15,  and  the  sum  of  their  squares  is  83.     Find  the 

numbers. 

Let  x  —  y,  X,  X  +  y  represent  the  numbers. 

17.  Arithmetical  means  are  inserted  between  5  and  23,  so 

that  the  sum  of  the  first  two  is  to  the  sum  of  the  last 
two  as  2  is  to  5.     How  many  means  are  inserted  ? 

18.  Find  three  numbers  of  an  arithmetical  series  whose  sum 

shall  be  21,  and  the'sum  of  the  first  and  second  shall 
be  f  of  the  sum  of  the  second  and  third. 

19.  Find  three  numbers  whose  common  difference  is  1,  such 

that  the  product  of  the  second  and  third  exceeds  that 
of  the  first  and  second  by  ?. 

20.  How  many  terms  of  the  series  1,  4,  7 must  be  taken, 

in  order  that  the  sum  of  the  first  half  may  bear  to 
the  sum  of  the  second  half  the  ratio  10  :  31  ? 

21.  A  travels  uniformly  20  miles   a  day  ;    B  starts  three 

days  later,  and  travels  8  miles  the  first  day,  12  the 
second,  and  so  on,  in  arithmetical  progression.  In 
how  many  days  will  B  overtake  A  ? 


SERIES.  307 


22.  A  number  consists  of  three  digits  which  are  in  arith- 

metical progression  ;  and  this  number  divided  by  the 
sum  of  its  digits  is  equal  to  26  ;  but  if  198  be  added 
to  it,  the  digits  in  the  units'  and  hundreds'  places  will 
be  interchanged.     Required  the  number. 

23.  The  sum  of  the  squares  of  the  extremes  of  four  numbers 

in  arithmetical  progression  is  200,  and  the  sum  of  the 
squares  of  the  means  is  136.    What  are  the  numbers? 

24.  Show  that  if  any  even  number  of  terms  of  the  series  1, 

3,  5 be  taken,  the  sum  of  the  first  half  is  to  the 

sum  of  the  second  half  in  the  ratio  1  :  3. 

25.  A  and  B  set  out  at  the  same  time  to  meet  each  other 

from  two  places  343  miles  apart.  Their  daily  jour- 
neys are  in  arithmetical  progression,  A's  increase 
being  2  miles  each  day,  and  B's  decrease  being  5 
miles  each  day.  On  the  day  at  the  end  of  which 
they  met,  each  travelled  exactly  20  miles.  Find  the 
duration  of  the  journey. 

26.  Suppose  that  a  body  falls  through  a  space  of  16y^j  feet  in 

the  first  second  of  its  fall,  and  in  each  succeeding  sec- 
ond 32J  more  than  in  the  next  preceding  one.  How 
far  will  a  body  fall  in  20  seconds  ? 

27.  The  sum  of  five  numbers  in  arithmetical  progression  is 

45,  and  the  product  of  the  first  and  fifth  is  |  of  the 
the  product  of  the  second  and  fourth.  Find  the 
numbers. 

28.  If  a  full  car  descending  an  incline  draw  up  an  empty 

one  at  the  rate  of  Ij  feet  the  first  second,  4  J  feet  the 
next  second,  7i  feet  the  third,  and  so  on,  how  long 
will  it  take  to  descend  an  incline  150  feet  in  length  ? 
What  part  of  the  distance  will  the  car  have  descended 
in  the  first  half  of  the  time  ? 


308  ALGEBRA. 

Geometrical  Series. 

384.  A  series  is  called  a  G-eometrical  Series  or  a  Geomet- 
rical Progressioii  when  each  succeeding  term  is  obtained  by 
multiplying  the  preceding  term  by  a  constant  multipUer. 

385.  The  general  representative  of  such  a  series  will  be 

a,  ar,  ar^,  ar^,  ar^ , 

in  which  a  is  the  first  term  and  r  the  constant  multiplier  or 
ratio. 

386.  Since  the  exponent  of  r  increases  by  1  for  every 
term,  the  exponent  will  always  be  1  less  than  the  number 
of  the  term  ;  so  that  the 

nth  term  =  ar"~^. 

387.  If  the  nth  term  be  denoted  by  I,  this  equation  be- 
'=°'^''  l  =  ar-\  (1) 

388.  The  geometrical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
a  geometrical  series. 

389.  If  a  and  b  denote  two  numbers,  and  G  their  geo- 
metrical mean,  then,  by  definition  of  a  geometrical  series, 

a    a' 

.'.  G  =  ^ab.  (2) 

390.  Sometimes  it  is  required  to  insert  several  geometri- 
cal means  between  two  numbers. 


SERIES.  309 

If  772,  =  the  number  of  means,  then  7n-\-2  =  n,  the  whole 
number  of  terms ;  and  if  m  +  2  be  substituted  for  n  in  the 
equation  l=ar^-\ 

the  result  is  1  =  ar~+^, 

.^  ^+1  ^  [  (3) 

a 
Thus,  if  it  be  required  to  insert  three  geometrical  means  between 
3  and  48,  the  value  of  r  is  found  to  be 

,-  =  48  =  16. 
3 
.'.  r  =  2, 
and  the  series  will  be         3,  6,  12,  24,  48. 

391.  If  I  denote  the  last  term,  a  the  first  term,  n  the 
number  of  terms,  r  the  common  ratio,  and  s  the  sum  of  the 
n  terms,  then 

s  =  a -{- ar -}- ai-^ -{- ar^ -\- +  ar*"'^. 

Multiply  by  r,  rs  ^=  ar -\- ar^ -}- a?^ -{- +  a?***~^  +  o^^- 

Therefore,  by  subtracting  the  first  equation  from  the 
second,  n 

or  (r  —  1)  s  =  a  (r"  —  1), 

.■.a  =  "(^"-l).  (4) 

r—  1 

392.  When  r  is  <  1,  this  formula  will  be  more  convenient 
if  written 

,_a(l-r-)^ 
1  — r 

393.  Since  /=ar"-\ 

rl  =  ar^y 

and  (4)  may  be  written      s  =  '^- ^. 


In  working  out  the  following  results,  the  student  will  make  use 

of  the  two  equations,  l=-ar^-'^  and  s  =  ^ — ^^ . 

r  —  \ 


310 


ALGEBRA. 


No. 


Given. 


a  r 
a  n 
r  n 


Required. 


Results. 


g  +  (r  - 1)  s 
r 
Z(s_^)«-i_a(s-a)«-i-0. 
,  _  (r  —  1)  sr'^-^ 


a  r  n 
a  r  Z 
a  n  I 
r  n  I 


a  (r^  —  1) 

r-1 
rl  —  a       _^ 
r-1' 

yM  .^W — 1 


9 

10 
11 
12 


r  71  ( 
r  n  s 
r  I  s 
n  I  s 


^n— 1 

r«-l 
a  =  rZ  —  (r  —  1)  s. 

a  (s  _  a)n-l  _  z  (s  _  ^)n-i  =  0. 


13 
14 
15 
16 
17 
18 
19 
20 


a  n  I 

a  n  s 

a  I  s 

n  I  s 


»  n. 


s         s  —  a^ 

r" r  -i =  0. 

a  a 


ifTi  . 


-I 


s-l 


=  0. 


a  r  I 

a  r  s 

a  I  s 

r  I  » 


log  r 

_  log  [a  +  {r—  1)  s]  -  log  a 


logr 
log  I  —  log  a 


+  1. 


log  {s-a)-  log  (s  -  I) 
^  _  log  ^  -  log  [Ir  -  (r  -  1)  g]  ^  .^^ 
log  r 


SERIES. 


311 


Exercise  CXVII. 

1.  Find  the  seventh  term  of  2,  6,  18 

sixth  term  of  3,  6,  12 

ninth  term  of  6,  3,  1^ 

eighth  term  of  1,  —  2,  4 

twelfth  term  of  a?,  x*,  a^ 

fifth  term  of  4  a,  —  6ma^,  drri^a^ 

2.  Find  the  geometrical  mean  between  18a;^y  and  ^Oxy^z. 

3.  ^ind  the  ratio  when  the  first  and  third  terms  are  5  and 

80  respectively. 

4.  Insert  two  geometrical  means  between  8  and  125  ;   and 

three  between  14  and  224. 

5.  If  a  =  2  and  r  =  3,  which  term  will  be  equal  to  162  ? 

6.  The  fifth  term  of  a  geometrical  series  is  48,  and  the  ratio 

2.     Find  the  first  and  seventh  terms. 

7.  Find  the  sum  of 

3  +  6  +  12  + to  eight  terms. 

1  —  3  +  9  — to  seven  terms. 

8  +  4  +  2  + to  ten  terms. 

.1  +  .5  +  2.5  +.....  to  seven  terms. 

m r  +  T^— to  live  terms. 

4      16 

8.  The  population  of  a  city  increases  in  four  years  from 

10,000  to  14,641.     What  is  the  rate  of  increase  ? 

9.  The  sum  of  four  numbers  in  geometrical  progression  is 

200,  and  the  first  term  is  5.     Find  the  ratio. 

10.   Find  the  sum  of  eight  terms  of  a  series  whose  last  term 
is  1,  and  fifth  term  i. 


312  ALGEBRA. 


11.  In  an  odd  number  of  terms,  show  that  the  product  of 

the  first  and  last  will  be  equal  to  the  square  of  the 
middle  term. 

12.  The  product  of  four  terms  of  a  geometrical  series  is  4, 

and  the  fourth  term  is  4.     Determine  the  series. 

13.  If  from  a  line  one-third  be  cut  ofif,  then  one-third  of  the 

remainder,  and  so  on,  what  fraction  of  the  whole  will 
remain  when  this  has  been  done  five  times  ? 

14.  Of  three  numbers  in  geometrical  progression,  the  sum 

of  the  first  and  second  exceeds  the  first  by  3,  and  the 
sum  of  the  first  and  third  exceeds  the  second  by,  21. 
"What  are  the  numbers  ? 

15.  Find  two  numbers  Avhose  sum  is  3^^  and  geometrical 

mean  1^? 

16.  A  glass  of  wine  is  taken  from  a  decanter  that  holds  ten 

glasses,  and  a  glass  of  water  poured  in.  When  this 
has  been  done  five  times,  what  part  of  the  contents  is 
wine? 

17.  A  vessel  contains  16  gallons  of  brandy  and  2  of  water. 

If  li  gallons  be  drawn  off  each  day,  and  the  same 
amount  of  brandy  be  poured  in,  how  much  brandy 
will  there  be  in  the  vessel  at  the  end  of  four  days  ? 

18.  Find  three  numbers  in  geometrical  progression  whose 

sum  is  13  and  the  sum  of  their  squares  91. 

19.  The  difference  between  two  numbers  is  48,  and  the  arith- 

metical mean  exceeds  the  geometrical  by  18.  Find 
the  numbers. 

20.  There  are  four  numbers  in  geometrical  progression,  the 

second  of  which  is  less  than  the  fourth  by  24,  and  the 
sum  of  the  extremes  is  to  the  sum  of  the  means  as  7 
to  3.     Find  the  numbers. 


SERIES.  313 

21.  A  number  consists  of  three  digits  in  geometrical  pro- 
gression. The  sum  of  the  digits  is  13 ;  and  if  792  be 
added  to  the  number,  the  digits  in  the  units'  and 
hundreds'  places  will  be  interchanged.  Find  the 
number. 

394.  When  r<  1,  a  geometrical  series  has  its  terms  con- 
tinually decreasing ;  and  by  increasing  n,  the  value  of  the 
nth  term,  ar"~^  may  be  made  as  small  as  we  please,  though 
not  absolutely  zero. 

395.  The  formula  for  the  sum  of  a  terms, 

g  (1  -  r**) 
1-r 

may  be  written . 

1  —  r      1  —  r 

By  increasing  n  indefinitely,  the  value  of becomes 

1  —  r 

indefinitely  small,  so  that  the  sum  of  n  terms  approaches 
indefinitely  to as  its  limit. 

Ex.  Find  the  limit  of  1  -  J  +  }  -  \ 

Here  a  =  l,  andr  =  — ^, 

_a___J____l 
1-r      l-(-^)      1  +  ^      3 


and  therefore  the  limit      ^     = = =  -.  Am. 


22.    Find  the  limits  of  the  sums  of  the  following  infinite 
series : 

4  +  2  +  1-1- 2-li  +  l- 

i  +  i  +  *+ .1  +  . 01 +  .001+ 

i-A+eV- 868686 

1~I  +  A- 54444 

i  +  T^  +  TV+ 83636 


314  ALGEBRA. 


*Hae,monical  Series. 

396.  A  series  is  called  a  Harmonical  Series,  or  a  Harmon- 
ical  Progression,  when  the  reciprocals  of  its  terms  form  an 
arithmetical  series. 

Hence,  the  general  representative  of  such  a  series  will  be 

1^    _1_^         1        1 

a     a-\-d     a -{-2d  a-\-(n—l)d 

397.  Questions  relating  to  harmonical  series  should  be 
solved  by  writing  the  reciprocals  of  its  terms  so  as  to  form 
an  arithmetical  series. 

398.  If  a  and  h  denote  two  numbers,  and  JI  their  har- 
monical mean,  then,  by  the  definition  of  a  harmonical  series, 

IT     a     I     H' 

•    2  _1   ,  \_a-\-h  , 

H     a      b        ah    ^ 

.    jr       2  ah 


a-{-h 


399.  Sometimes  it  is  required  to  insert  several  harmoni- 
cal means  between  two  numbers. 

Ex.  Let  it  be  required  to  insert  three  harmonical  means 
between  3  and  18. 

Find  the  three  arithmetical  means  between  ^  and  y^-. 

These  are  found  to  be  ^|,  ^f ,  y\ ;  therefore,  the  harmonical  means 
areH.H.-V-;  or  S^f,  5^,  8. 

*  A  harmonical  series  is  so  called  because  musical  strings  of  uniform 
thickness  and  tension  produce  harmony  when  their  lengths  are  represented 
by  the  reciprocals  of  the  natural  series  of  numbers ;  that  is,  by  the  series, 
1,  i,  h  i,  i,  etc. 


SERIES.  315 


Exercise  CXVIII. 

1.  Insert  four  harmonical  means  between  2  and  12. 

2.  Find  two  numbers  whose  difference  is  8  and  the  har- 

monical mean  between  them  1|. 

3.  Find  the  seventh  term  of  the  harmonical  series  3,  3-f-, 

4 

4.  Continue  to  two  terms  each  way  the  harmonical  series 

two  consecutive  terms  of  which  are  15,  16. 

5.  The  first  two  terms  of  a  harmonical  series  are  5  and  6. 

Which  term  will  equal  30  ? 

6.  The  fifth  and  ninth  terms  of  a  harmonical  series  are  8 

and  12.     Find  the  first  four  terms. 

7.  The  difference  between  the  arithmetical  and  harmonical 

means  between  two  numbers  is  1|-,  and  one  of  the 
numbers  is  four  times  the  other.    Find  the  numbers. 

8.  Find   the    arithmetical,   geometrical,    and    harmonical 

means  between  two  numbers  a  and  b ;  and  show  that 
the  geometrical  mean  is  a  mean  proportional  between 
the  arithmetical  and  harmonical  means.  Also,  ar- 
range these  means  in  order  of  magnitude. 

9.  The  arithmetical  mean  between  two  numbers  exceeds 

the  geometrical  by  13,  and  the  geometrical  exceeds 
the  harmonical  by  12.     What  are  the  numbers  ? 

10.  The  sum  of  three  terms  of  a  harmonical  series  is  11,  and 

the  sum  of  their  squares  is  49.     Find  the  numbers. 

11.  When  a,  b,  c  are  in  harmonical  progression,  show  that 

a  :  c  :  :  a  —  b  :  b  —  c. 


CHAPTER    XXII. 

Choice* 

400.  If  three  paths,  A,  JB,  and  C,  lead  to  the  top  of  a 
mountain,  there  is  obviously  a  choice  of  three  different  ways 
of  ascending  the  mountain  ;  and  when  the  top  of  the  moun- 
tain is  reached,  there  is  again  a  choice  of  three  different 
ways  of  descending. 

How  many  different  ways  are  there  of  doing  both  ? 

If  a  traveller  ascend  by  ^,  he  may  descend  by  ^,  ^,  or 
C.  This  makes  three  ways  of  doing  both.  If  he  ascend  by 
_S,  he  may  descend  by  A,  B,  or  C;  and  again,  if  he  ascend 
by  C,  he  may  descend  by  ^,  ^,  or  O. 

Therefore,  there  are  3  X  3  =  9  ways  in  all  of  doing  both. 

These  ways  may  be  indicated  as  follows : 


1. 

A  and  A. 

4. 

B  and  A. 

7. 

Cand^. 

2. 

A  and  B. 

5. 

B  and  B. 

8. 

(7and^. 

3. 

A  and  a 

6. 

B  and  a 

9. 

(7  and  a 

401.  Suppose  the  traveller  does  not  wish  to  ascend  and 
descend  by  the  same  path,  then  what  choice  has  he  ? 

He  has  a  choice  of  three  different  ways  in  ascending. 

But  he  has  a  choice  of  only  two  ways  in  descending. 

If  each  of  the  three  ways  of  ascending  be  joined  to  the 
two  eligible  ways  of  descending,  the  result  is  3x2  =  6 
ways  of  doing  both. 

This  chapter  is  based  upon  Whitworth's  Choice  and  Chance,  and  many 
of  the  examples  have  been  taken  from  that  elegant  work. 


CHOICE. 

317 

These  ways  are : 

1.  A  and  B. 

2.  A  and  C. 

3.  B  and  A. 

4.  B  and  C. 

5.  Cand^. 

6.  Ce^ndB. 

402.  If  a  box  contain  five  capital  letters,  A,  B,  C,  D,  E, 
and  three  small  letters,  x,  y,  z,  in  how  many  different  ways 
may  a  capital  letter  and  a  small  letter  be  selected  ? 

A  capital  letter  may  be  selected  in  5  ways. 

With  each  capital  letter  selected,  a  small  letter  can  be 
joined  in  3  ways.  So  that  the  number  of  different  ways  in 
which  the  selection   can  be  made   is  3  X  5  =  15.     These 


vays  are  : 

Ax 

Bx 

Cx 

Dx 

Ex 

Ay 

Az 

By 
Bz 

Oy 

Cz 

By 
Dz 

Ey 
Ez 

403.    Hence  the  fundamental  principle  of  choice  : 
I.    If  one  thing  can  he  done  in  a  different  ways,  and  {when 
it  has  been  done  in  any  one  of  these  ways)  another  thing  can 
be  don£  in  b  different  ways,  then  both  can  be  done  in  a  X  b  dif- 
ferent ways. 

For,  corresponding  to  each  of  the  a  ways  of  doing  the  first  thing, 
there  are  h  ways  of  doing  the  second  thing.  Therefore,  altogether, 
there  are  axh  ways  of  doing  both  things. 

(1)  On  a  shelf  are  7  English  and  5  French  books.     In  how 

many  ways  can  one  of  each  be  chosen  ? 
7x5  =  35.  Ans. 

(2)  On  a  shelf  are  7  English,  5  French,  and  9  German 

books.     In  how  many  ways  can  two  books  be  chosen 

so  that  they  shall  be  in  different  languages  ? 

An  EngUsh  book  and  a  French  book  can  be  chosen  in  7  X  5 
=  35  ways.  A  French  book  and  a  German  book  in  5  x  9  =  45 
ways.    An  English  book  and  a  German  book  in  7  X  9  =  63  ways. 

Hence,  there  is  a  choice  of  35  +  45  +  63  =  143  ways.  Ans. 


318  ALGEBRA. 


(3)  Out  of  8  different  pairs  of  gloves,  in  how  many  differ- 

ent ways  can  a  right-hand  and  a  left-hand  glove  be 

chosen  which  shall  not  form  a  pair  ? 

A  right-hand  glove  can  be  chosen  in  8  ways ;  and  when  it 
is  chosen  there  are  7  left-hand  gloves,  any  one  of  which  may 
be  put  with  it  without  making  a  pair.  Hence,  the  choice  is  in 
8  X  7  =  56  ways. 

(4)  Out  of  the  ten  digits,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  how  many 

numbers  each  consisting  of  two  figures  can  be  formed  ? 

Since  0  has  no  value  in  the  left-hand  place,  the  left-hand  place 
can  be  filled  in  9  ways. 

The  right-hand  place  can  be  filled  in  10  ways,  since  repetitions 
of  the  digits  are  allowed  (as  22,  33,  etc.). 

Hence,  the  whole  number  is  9  X  10  =  90. 

(5)  How  many  odd  numbers  consisting  of  two  figures  can 

be  formed  with  the  ten  digits  ? 

The  left-hand  place  can  be  filled  in  9  ways ;  the  right-hand 
place  in  only  5  ways,  since  it  must  be  either  1,  3,  5,  7,  or  9. 
Hence,  the  number  is,  9  x  5=  45.   Ans. 

404.    By  a  simple  extension  of  Rule  I.  it  is  evident  that, 

II.  If  one  thing  can  he  done  in  a,  ways,  and  then  a  second 
thing  can  he  done  in  b  ways,  then  a  third  in  c  ways,  then  a 
fourth  in  d  ways,  etc.,  the  numher  of  ways  of  doing  all  the 
things  will  ^e  a  X  b  X  c  X  d,  etc. 

(1)    In  how  many  ways  can  four  Christmas  presents  be  given 

to  four  boys,  one  to  each  boy  ? 

The  first  present  may  be  given  to  any  one  of  the  boys ;  hence 
there  are  4  ways  of  disposing  of  it. 

The  second  present  may  be  given  to  any  one  of  the  other  three 
boys  ;  hence  there  are  3  ways  "of  disposing  of  it. 

The  third  present  may  be  given  to  either  of  the  other  two 
boys ;  hence  there  are  2  ways  of  disposing  of  it. 


CHOICE.  319 

The  fourth  present  must  be  given  k>  the  last  boy ;  hence  there 
is  only  1  way  of  disposing  of  it. 

There  are,  then,  4x3x2x1  =  24  ways. 

(2)  In  how  many  ways  can  five  presents  be  given  to  two 

children  ? 

Each  present  may  be  disposed  of  in  2  ways ;  for  it  may  be  given 
to  either  child.  Hence,  the  whole  number  of  ways  of  giving  the 
presents  is  2x2x2x2x2  =  32. 

(3)  In  how  many  ways  can  five  presents  be  divided  between 

two  children? 

This  question  differs  from  the  last  only  in  the  fact  that  a  divi- 
sion of  the  gifts  excludes  the  two  ways  in  which  either  child 
receives  all  the  gifts. 

Hence,  there  are  32  —  2  =  30  ways. 

(4)  In  how  many  ways  can  x  things  be  given  to  n  persons? 

71*.  Arts. 

(5)  In  how  many  ways  can  a  vowel  and  a  consonant  be 

chosen  out  of  the  alphabet  ? 

Since  there  are  in  the  alphabet  6  vowels  and  20  cotisonants,  a 
vowel  can  be  chosen  in  6  ways  and  a  consonant  in  20  ways,  and 
both  (Rule  I.)  in  6  x  20  =  120  ways. 

(6)  In  how  many  ways  can  a  two-lettered  word  be  made, 

containing  one  vowel  and  one  consonant  ? 

The  vowel  can  be  chosen  in  6  ways  and  the  consonant  in  20 
ways ;  and  then  each  combination  of  a  vowel  and  a  consonant 
can  be  written  in  2  ways  ;  as  ac,  ca. 

Hence,  the  whole  number  of  ways  is  6  X  20  X  2  =  240. 

405.  The  last  two  examples  show  the  difference  between 
a  selection  or  combination  of  different  things,  and  an  arrange- 
ment or  permutation  of  the  same  things. 


320  ALGEBRA. 


Thus,  ac  form  a  combination  of  a  vowel  and  a  consonant,  and  ac 
and  ca  form  two  different  arrangements  of  this  combination. 

From  (5)  it  is  seen  that  120  different  combinations  can  be  made 
with  a  vowel  and  a  consonant ;  and  from  (6)  it  is  seen  that  240  differ- 
ent permutations  can  be  made  with  the  same. 

Again,  a,  &,  c  is  a  selection  of  three  letters  from  the  alphabet.  This 
selection  then  admits  of  6  different  arrangements  or  permutations,  as 
follows : 

abc  bca  cab 

acb  bac  cba 

406.  A  selection  or  combination  of  any  number  of  ele- 
ments or  things,  means  a  group  of  that  number  of  elements 
or  things  put  together  without  regard  to  their  order  of 
sequence.  An  arrangement  or  permutation  of  any  number 
of  elements  or  things  means  a  group  of  that  number  of  ele- 
ments or  things  put  together  with  reference  to  their  order 
of  sequence. 

Arrangements  or  Permutations. 

407.  In  how  many  ways  can  the  letters  of  the  word 
Cambridge  be  arranged,  taken  all  at  a  time? 

There  are  nine  letters.  In  making  any  arrangement  any  one  of 
the  letters  may  be  put  in  the  first  place.  Hence,  the  first  place  can 
be  filled  in  9  ways.  Then  the  second  place  can  be  filled  with  any 
one  of  the  remaining  eight  letters  ;  that  is,  in  8  ways. 

In  like  manner,  the  third  place  in  7  ways,  the  fourth  place  in  6 
ways,  and  so  on ;  and,  lastly,  the  ninth  place  in  1  way. 

If  the  nine  places  be  indicated  by  Roman  numerals,  the  result 
(Rule  II.)  is  as  follows : 

I.  II.  III.  IV.  V.  VI.  VII.  VIII.  IX. 
9x8x7x6x5x4x  3   X    2   X    1  =  362,880  ways. 

408.  Hence,  it  will  be  seen  that, 

III.    The  nuTnber  of  arrangements  or  permutations  ofn 
different  elements  or  things  tahen  all  at  a  time  is 
n(n-l)(n-2)(n-d) 3x2x1. 


CHOICE.  321 


For,  the  first  place  may  be  filled  in  n  ways,  then  the  second  place 
in  n  —  1  ways,  then  the  third  place  in  n  —  2  ways,  and  so  on  to  the 
last  place,  which  can  be  filled  in  only  1  way. 

Hence  (Rule  II.)  the  whole  number  of  arrangements  is  the  con- 
tinued product  of  all  these  numbers, 

n(n-])(n-2)(n-3) 3x2x1. 

409.  For  the  sake  of  brevity  this  product  is  written  \n, 
and  is  read  factorial  n. 


410.    It  will  also  be  evident  that, 

IV,  The  number  of  arrangements  of  n  different  elements, 
taken  T  at  a  time,  is 

n(n  —  l)(n  —  2) to  r  factors, 

thai  is,  n(n  —  l)(n  —  2) [n  —  (r  —  1)], 

or  n(h  —  l)(n—2) (n  — r-fl). 

For  the  first  place  can  be  filled  in  n  ways,  the  second  in  n  —  1 
ways,  the  third  place  in  n  —  2  ways,  and  the  rth  place  in  n  —  (r  —  1) 
ways. 

(1)  A  shelf  contains  4  English  books,  5  French  books,  and 

6  German  books ;  in  how  many  ways  can  these  books 

be  arranged? 

[15  =  1,307,674,368,000  ways. 

(2)  In  how  many  ways  can  these  books  be  arranged,  if  the 

books  of  each  language  be  kept  together  ? 

The  English  books  can  be  arranged  (Rule  III.)  in  [4  ways,  the 
French  books  in  [5  ways,  and  the  German  books  in  [6  ways. 
Also,  the  three  set?  can  be  arranged  in  [3  different  orders.  Hence, 
the  number  of  ways  in  which  the  whole  can  be  done  is  (Rule  II.) 

[4x[5x[6xt3  =  12,411,600. 


322  ALGEBRA. 

(3)  Of  the  arrangements  possible  with  the  letters  of  the 

word  Cambridge  : 

(i.)    How  many  will  begin  Avith  a  vowel  ? 
(ii.)    How  many  will  both  begin  and  end  with  a  vowel  ? 

In  fining  the  nine  places  of  any  arrangement  in  case  (i.),  tiie 
first  place  can  be  filled  in  only  3  ways,  the  other  places  in  [8 
ways. 

In  case  (ii.)  the  first  place  can  be  filled  in  3  ways,  the  last  place 
in  2  ways  (one  vowel  having  been  used),  and  the  remaining  seven 
places  in  [7  ways. 

Hence,  the  answer  to  (i.)  is  3  x  [8  ==  120,960 ; 

to  (ii.)  is  3  X  2  X  [7  =  30,240. 

(4)  With  the  letters  of  the  word  Cambridge^  how  many 

arrangements  can  be  made  : 
(i.)    Each  beginning  with  the  word  Gmn  ? 

(ii.)    Keeping  the  letters  Cam  always  together? 

For  case  (i.)  the  answer  is  evidently  [6  ;  since  our  only  choice 
lies  in  arranging  the  remaining  six  letters  of  the  word. 

Case  (ii.)  may  be  resolved  into  arranging  Cam  and  the  last  six 
letters,  regarded  as  seven  distinct  elements,  and  then  arranging 
the  letters  Cam. 

The  first  can  be  done  in  |_7  ways,  and  the  second  in  [3  ways. 
Hence  (Eule  II.)  both  can  be  done  in  [7  X  [3  =  30,240  ways. 

(5)  In  how  many  ways  can  the  letters  of  the  word  Cam- 

bridge be  written : 
(i.)    Without  changing  the  place  of  any  vowel  ? 
(ii.)   Without  changing  the  order  of  any  vowel  ? 
(iii.)   Without  changing  the  relative  order  of  vowels 

and  consonants  ? 
In  (i.)  the  second,  sixth,  and  ninth  places  can  be  filled  each  in 
only  1  way ;  the  other  places  in  [6  ways. 

Therefore,  the  whole  number  of  ways  is  [6  =  720. 
In  (ii.)  the  vowels  in  the  different  arrangements  are  always 
kept  in  the  order  a,  i,  e.    One  of  the  six  consonants  can  be  placed 
in  4  ways :  hefore  a,  between  a  and  i,  between  i  and  e,  and  after  e. 


CHOICE.  323 


Then  a  second  consonant  can  be  placed  in  5  ways,  a  third 
consonant  in  6  ways,  a  fourth  consonant  in  7  ways,  a  fifth  con- 
sonant in  8  ways,  and  the  last  consonant  in  9  ways.  Hence 
(Rule  II.)  the  whole  number  of  ways  is  4x5x6x7x8x9 
=  60,480. 

In  (iii.)  the  vowels  can  be  arranged  in  [3  ways,  and  the  conso- 
nants in  [6  ways.  Hence  (Rule  II.)  the  number  of  ways  is 
[3  X  |6  =  4,320' 

(6)  In  how  many  ways  can  4  persons,  A,  B,  C,  D,  sit  at  a 

round  table? 

If  the  four  places  are  not  regarded  as  relative  to  each  other, 
then  the  whole  number  of  ways  is  [4  =  24.  But  if  the  four  places 
are  regarded  as  relative  to  each  other,  then  by  placing  one  as  A, 
in  one  position,  and  by  arranging  the  others  in  the  other  three 
positions,  the  whole  number  of  ways  is  [3  =  6. 

(7)  In  how  many  ways  can  6  persons  form  a  ring? 

Here  relative  position  is  required.  Hence,  the  whole  number 
of  ways  is  [5  =  120. 

(8)  How  many  three-lettered  words  can  be  made  from  the 

alphabet,  no  letter  being  repeated  in  the  same  word  ? 
26  X  25  X  24  =  15,600.   Am. 

(9)  How  many  four-lettered  words? 

26  X  25  X  24  X  23  =  358,800.  Ans. 

(10)  How  many  different  arrangements  can  be  made  of  the 

letters  in  the  word  a/e^ 

Distinguish  the  e's  thus,  e^,  e^,  and  arrange  as  follows : 

^  y^t        «i  ^.  3/        y  «t «. 
«,  y  «i        «.  «i  y        y «,  «i 

These  six  arrangements  become  three  when  the  e's  are  not  dis- 
tinguished. That  is,  each  pair  of  arrangements  produced  by 
permuting  the  e's  is  reduced  to  a  single  arrangement. 

Hence,  the  number  of  different  arrangements  is  found  by  di- 
viding [3  by  [2 ,  that  is,  by  dividing  the  number  of  arrangements 
possible  when  the  letters  are  all  different  by  the  number  of  ways 
in  which  the  two  e's  can  be  permuted. 


324  ALGEBRA. 


411.  In  how  many  different  orders  can  the  letters  a,  a,  a, 
X,  y  be  written  ? 

If  the  letters  were  all  different,  the  answer  (Rule  III.)  would  be  [5; 
but  the  three  cCs,  may  be  permuted  in  |^  =  6  ways. 

Hence,  the  [5  arrangements  may  be  divided  into  six  groups,  each 
group  constituting  but  a  single  arrangement  of  the  given  letters. 

Hence,  the  whole  number  of  given  orders  is  —  =  20. 

By  the  same  course  of  reasoning,  the  whole  number  of 
different  orders  of  a,  a,  a,  x,  x  is  found  to  be 


^    -10. 


[3(2 


412.    In  like  manner  for  any  other  numbers.     Hence, 
V.    The  nuTTiher  of  arrangements  of  n  elements,  of  which 
p  are  alike,  q  others  are  alike,  andi  others  are  alike ,is 


(1)    In  how  many  ways  can  the  letters  of  the  word  Missis- 
sippi be  arranged  ? 
Ill 


[4  [4  [2 


34,560.  Ans. 


(2)  In  how  many  different  orders  can  a  row  of  4  white 

balls  and  3  black  balls  be  arranged  ? 

-^=^-35.  Ans. 

L4|_3 

(3)  In  how  many  ways  can  4  white  balls  and  3  black  balls 

.be  placed  in  a  row,  if  the  balls  are  all  different  in. 
size. 
[7  =  5040.  Ans. 


CHOICE.  325 


413.  In  case  the  n  elements  to  be  arranged  are  all  differ- 
ent, but  repetitions  of  them  are  allowed,  then  in  making  any 
arrangement,  every  place  to  be  filled  can  be  filled  in  n  ways, 
since  we  may  always  repeat  an  element  already  used. 
Hence,  corresponding  to  Rules  III.  and  IV.,  the  following 
rules  apply  to  cases  in  which  repetitions  are  allowed  : 

VL  The  number  of  arrangements  of  n  different  ele^ments, 
taken  all  at  a  time,  when  repetitions  are  allowed,  is  n**. 

VII.  The  number  of  arrangements  of  n  different  elements, 
taken  I  at  a  time,  when  repetitions  are  allowed,  is  n'. 

(1)  How  many  three-lettered  words  can  be  made  from  the 

alphabet,  when  repetitions  are  allowed  ? 

26^  =  17,576.  Ana. 

(2)  How  many  three-lettered  words  can  be  made  from  the 

6  vowels  when  repetitions  are  allowed  ? 

6'  =  216.  Am. 

(3)  A  railway  signal  has  3  arms,  and  each  arm  may  take 

four  different  positions,  including  the  position  of  rest. 
How  many  signals  in  all  can  be  made  ? 

43  -  1  =  63.  Ans. 

(4)  In  the  common  system  of  notation,  how  many  numbers 

can  be  formed  consisting  of  not  more  than  5  digits  ? 

All  the  possible  numbers  may  be  regarded  as  consisting  of 
each  5  digits,  by  prefixing  zeros  to  the  numbers  consisting  of  less 
than  5  digits.     Thus,  247  may  be  written  00247. 

Hence,  every  possible  arrangement  of  5  digits  out  of  the  10 
digits  will  give  one  of  the  required  numbers  except  00000 ;  and 
the  answer  is  10*  —  1  =  99999  ;  that  is,  all  the  numbers  between 
0  and  100,000. 


326  ALGEBRA. 


(5)  With  the  digits  0,  1,  2,  3,  4,  5,  how  many  numbers  be- 
tween 1000  and  4000  can  be  formed  ? 

Here  we  have  to  fill,  in  each  possible  case,  four  places.  The 
first  place  can  be  filled  with  1,  2,  or  3,  that  is,  in  3  ways ;  the 
second,  third,  and  fourth  places  each  in  6  ways.  Therefore,  the 
answer  is  3  x  6^  =  648. 

(6)  With  the  same  digits,  0,  1,  2,  3,  4,  5,  and  between  the 

same  limits,  1000  and  4000: 
(i.)    How  many  even  numbers  can  be  formed  ? 
(ii.)    How  many  odd  numbers  can  be  formed  ? 

(iii.)    How  many  numbers  divisible  by  5  ? 

Evidently  (i.)  and  (ii.)  are  like  Ex.  (5),  except  that  the  last  place 
can  be  filled  in  only  3  ways.  In  (i.)  the  last  place  must  be  filled 
by  0,  2,  or  4 ;  in  (ii.)  the  last  place  must  be  filled  by  1,  3,  or  5. 

Hence,  the  answer  in  each  of  these  cases  is  3x6x6x3  =  324, 

In  (iii.)  the  last  digit  must  be  either  0  or  5 ;  and  the  answer 
for  this  case  is  3x6x6x2  =  216. 

Combinations. 

414.  In  how  many  ways  can  3  vowels  be  selected  from 
the  5  vowels  a,  e,  i,  o,  u. 

The  number  of  ways  in  which  we  can  arrange  3  vowels  out  of  5  is 
(Rule  IV.)  5  X  4  X  3  =  60. 

These  60  arrangements  might  be  obtained  by  first  forming  all  the 
possible  selections  of  the  3  vowels  out  of  5,  and  then  arranging  the  3 
vowels  in  each  selection  in  as  many  ways  as  possible. 

The  3  vowels  of  each  selection  may  be  arranged  in  [3  =  6  ways. 

Hence  (Rule  II.), 

Number  of  selections  X  6  =  number  of  arrangements  =  60. 

Therefore,  number  of  selections  =  ^  =  10. 

415.  In  general, 

VIII.  Out  of  n  different  elements,  the  number  of  selections 
of  r  elements  is  equal  to  the  numher  of  arrangements  of  n  ele- 
ments divided  hy  [r. 


CHOICE.  327 


For,  lot  s  =  number  of  ways  of  selecting  r  elements  out  of  n  ele- 
ments. Then  the  r  elements  thus  selected  may  be  arranged  (Rule  III.) 
in  [r_  different  ways.  Therefore  (Rule  I.)  s  X  [r  =  number  of  arrange- 
ments of  n  elements  taken  r  at  a  time. 

_  number  of  arrangements 

[r 

(By  Rule  IV.)   The  numerator  of  this  fraction  is  equal  to 

n{n-l){n-2) [n-(r-l)]. 

_n(n-l)(n-2) (n-r-\-l) 

If  both  terms  of  this  fraction  be  multiplied  by  \n~r, 

\n 
the  result  is  s  =  - — ; 

Lr  1^  ~  ^ 

If  n  —  r  =p,  then  n  =p  +  r,  and  this  formula  may  be  written 
\P  +  r 

416.  The  value  of  this  fraction  is  not  altered  if  p  and  r 
be  interchanged.     Hence, 

IX.  Out  o/"  p  4"  r  different  elements,  the  numher  of  ways 
in  which  p  elements  can  he  selected  is  the  same  as  the  num- 
her of  ways  in  which  r  elements  can  he  selected. 

Thus,  out  of  8  elements,  3  elements  can  be  selected  in  the  same  num- 
ber of  ways  as  5  elements ;  namely, 

^    -8-7.6      ,« 

(1)    Out  of  20  consonants,  in  how  many  ways  can  18  be 
selected  ? 

The  18  can  be  selected  in  the  same  number  of  ways  as  2 ;  and 
the  number  of  ways  in  which  2  can  be  selected  (Ruk  VIII.)  is 


328  ALGEBRA. 


(2)  In  how  many  ways  can  the  same  choice  be  made  so  as 

always  to  include  the  letter  B  ? 

Taking  B  first  we  must  then  select  17  out  of  the  remaining  19 

consonants.     This  can  be  done  in 

19x18      TK, 

- — ^ =  171  ways. 

(3)  In  how  many  ways  can  the  same  choice  be  made  so  as 

to  include  £  and  not  to  include  C? 

Taking  B  first,  we  have  then  to  choose  17  out  of  18,  C  being 
excluded.     This  can  be  done  in  18  ways. 

(4)  A  society  consists  of  50  members,  10  of  whom  are  phy- 

sicians.    In  how  many  ways  can  a  committee  of  6 
members  be  selected  so  as  to  include  1  physician. 
The  5  members  not  physicians  can  be  selected  in 

140 

ways, 

|_5   [35       -^  ' 

and  the  1  physician  in  10  ways.    Hence,  the  6  can  be  selected  in 

[40 
^  10  X     ~~    different  ways. 

o   I  oO 

(5)  In  how  many  ways  can  a  committee  of  6  members  be 

selected  so  as  to  include  at  least  one  physician? 
Six  members  can  be  selected  from  the  whole  society  in 

150 

Six  members  can  be  selected  from  the  whole  society,  so  as  to 
include  no  physician,  by  choosing  them  all  from  the  40  members 
who  are  not  physicians,  and  this  can  be  done  in 
[40 


[6   [34 


ways. 


150  [40  ,         ,  .     -,    .. 

Hence,  -^= ^=—  =  number  of  ways  of  selecting 

[6   [44     [6  [34 

the  committee  so  as  to  include  at  least- one  physician. 


CHOICE.  329 


(6) "  Out  of  20  Republicans  and  6  Democrats,  what  choice 

is  there  of  appointing  a  committee  consisting  of  3 

Republicans  and  2  Democrats  ? 

20  X  19  X  18 

The  Republicans  can  be  selected  in =  1140  ways ; 

n  \^  K  iX^Xo 

and  the  Democrats  in =  15  ways.     Hence,  the  whole  com- 

1X2 
mittee  can  be  appointed  in  1140  X  15  =  17,100  ways. 

(7)  From  20  Republicans  and  6  Democrats,  in  how  many 

ways  may  5  different  offices  be  filled,  three  of  which 
must  be  filled  by  Republicans,  and  the  other  two  by 
Democrats  ? 

The  first  three  offices  can  be  assigned  to  3  Republicans  in 
20  X  19  X  18  =  6840  ways  (Rule  II.) ;  and  the  other  two  offices 
can  be  assigned  to  2  Democrats  in  6  x  5  =  30  ways. 

There  is,  then,  a  choice  of  6840  x  30  =  205,200  different  ways. 

(8)  Out  of  20  consonants  and  6  vowels,  in  how  many  ways 

can  we  make  a  word  consisting  of  3  different  conso- 
nants and  2  different  vowels  ? 

20  X  19  X  18 

Three  consonants  can  be  selected  in — ~ =  1140  ways, 

6x5  1x2x3 

and  2  vowels  in  — :-^—  =  15  ways.     Hence  (Rule  I.)  the  5  letters 

can  be  selected  in  1140  x  15  =  17,100  ways. 

When  they  have  been  so  selected,  they  can  be  arranged  (Rule 
III.)  in  [5  =  120  different  orders.  Hence,  there  are  17,100  X 
120  =  2,052,000  different  ways  of  making  the  word. 

(d)  How  many  words  of  2  consonants  and  1  vowel  can  be 
formed  from  6  consonants  and  three  vowels,  the  vowel 
being  the  middle  letter  of  each  word  ? 

The  two  consonants  can  be  selected  in  15  ways ;  the  vowel  in 
3  ways.  Each  combination  of  the  2  consonants  and  1  vowel  can 
be  arranged  in  2x1x1  =  2  ways.  Hence,  the  number  of 
words  that  can  be  formed  is  15  X  3  X  2  =  90. 


330  ALGEBRA. 


(10)  How  many  words  of  3  consonants  and  3  vowels  can  be 
formed  from  the  alphabet,  if  one  of  the  vowels  is  to 
be  always  a  ? 

The  consonants  can  be  selected  in  — — — — —  =  1140  ways, 
A.x.  1    •     5x4      ..  1X2X3  -^  ' 

and  the  vowels  in  — ^-  =  10  ways, 
i  X  -^ 
Then  the  six  letters  of  each  combination  can  be  arranged  in 

1 6  =  720  ways.     Hence,  the  number  of  words  that  can  be  formed 

is  1140  X  10  X  720  =  8,208,000. 

417.  To  find  for  what  value  of  r  the  number  of  selections 
of  n  things,  tahen  x  at  a  time,  is  the  greatest. 

The  formula      ,  ^^  {n  -  \){n  -  2) (n  -  r  +  1) 

1  X2x3x r 

1  -ii.  n ^,  n  —  1  ^,n  —  2        ?i  —  r  +  1 

may  be  written      s  =  -  X X ■ — . 

^  1         2  3  r 

The  numerators  of  the  factors  on  the  right  side  of  this  equation 
begin  with  n,  and  form  a  descending  series  with  the  common  differ- 
ence 1;  and  the  denominators  begin  with  1,  and  form  an  ascending 
series  with  the  common  difference  1.  Therefore,  from  some  point  in 
the  series,  these  factors  become  less  than  1.  Hence,  the  maximum 
product  is  reached  when  that  product  includes  all  the  factors  greater 
than  1. 

When  n  is  an  odd  number,  the  numerator  and  the  denominator 

of  each  factor  will  be  alternately  both  odd  and  both  even ;  so  that 

the  factor  greater  than  1,  but  nearest  to  1,  will  be  the  factor  whose 

numerator  exceeds  the  denominator  by  2.     Hence,  in  this  case,  r  must 

have  such  a  value  that 

n  —  1 
n  —  r  +  l=r  +  2,      or      r  =  — ^ 

"When  n  is  an  even  number,  the  numerator  of  the  first  factor  will 
be  even  and  the  denominator  odd  ;  the  numerator  of  the  second  fac- 
tor will  be  odd  and  the  denominator  even ;  and  so  on,  alternately ; 
so  that  the  factor  greater  than  1,  but  nearest  to  1,  will  be  the  factor 
whose  numerator  exceeds  the  denominator  by  1.  Hence,  in  this  case, 
r  must  have  such  a  value  that 

n  — r  +  l  =  r  +  l,       or      r  =  — 


CHOICE.  331 


(1)  What  value  of  r  will  give  the  greatest  number  of  selec- 

tions out  of  7  things  ? 

Here  n  is  odd,  and  r  =  — '^^~  =  — ^^^^  =  3. 
2  2 

1x2x3 

Ifr==4,then  ,^  7  X  6  x  5  X  4^  3^^ 

1x2x3x4 

So  that,  when  the  number  of  things  is  odd,  there  will  be 
two  equal  numbers  of  selections ;  namely,  when  the  number 
of  things  taken  together  is  just  under  and  just  over  one-half 
of  the  whole  number. 

(2)  What  value  of  r  will  give  the  greatest  number  of  selec- 

tions out  of  8  things  ? 

Here  n  is  even,  and  r  =  -  =  -  =  4, 
'  2     2 

.^^8X7X6X5^^Q    ^ns. 
1x2x3x4 

So  that,  when  the  number  of  things  is  even,  the  number 
of  selections  will  be  greatest  when  one-half  of  the  whole 
are  taken  together. 

418.    It  may  be  shown  that, 

X.  The  number  of  ways  in  which  x  +  y  different  ele- 
ments can  be  divided  into  two  classes,  so  that  one  shall  con- 
tain X  and  the  other  j  elements,  is  equal  to  the  number  of 
ways  in  which  either  x  elements  or  y  elements  may  be  selected 
from  the  x  +  y  elements ;  or,. 

l^  +  y 

For  each  division  of  x  +  y  elements  into  two  classes,  one  consisting 
of  X  elements,  the  other  of  y  elements,  is  evidently  effected  by  making 
a  selection  of  x  elements  from  x  f  y  elements,  and  leaving  y  elements 
not  selected. 


332  ALGEBRA. 

In  like  manner  the  number  of  ways  in  wliicli  x-\-y-\-z 
different  elements  can  be  divided  into  3  classes,  containing 
X,  y,  and  z  elements  respectively,  is 

I    ~^  -Ij^  ;  and  so  on. 

\x\ii\z 

(1)  In  bow  many  ways  can  18  men  be  divided  into  2  classes 

of  6  and  12? 

[18 
[6  [12 

(2)  In  how  many  ways  can  18  men  be  divided  into  2  groups 

of  9  each  ? 

According  to  the  rule,  the  answer  would  be 

[18 
19[9 

The  two  groups,  considered  as  groups,  have  no  distinction ; 
therefore,  permuting  them  gives  no  new  arrangement,  and  the 
true  result  is  obtained  by  dividing  the  preceding  by  |_2,  and  is 

[2  [9  [9 

If  any  condition  be  added  that  shall  make  the  two  groups  dif- 
ferent, as,  if  one  group  wear  red  badges  and  the  other  blue,  then 
the  answer  would  be  n  g 

419.  Whenever  the  groups  are  indifferent,  Rule  IX.  gives 
each  arrangement  repeated  as  many  times  as  the  groups 
can  be  permuted  one  with  another  ;  that  is,  [2  when  there 
are  2  groups,  |^  when  there  are  3  groups,  and  so  on. 

Hence,  the  result  found  by  the  rule  must  be  divided  by 
\2,  [3,  etc.,  in  order  to  obtain  the  true  result. 


CHOICE.  333 


(1)  In  how  many  ways  can  the  52  cards  in  a  pack  be  di- 

vided among  4  players,  each  to  have  13  ? 

Here  the  assignment  of  each  group  to  a  different  player  makes 
the  groups  different;  and  the  answer  is 

[52 
[13  [13  [13  [is' 

(2)  In  how  many  ways  can  the  52  cards  of  a  pack  be  di- 

vided into  4  piles? 

Here  the  groups  are  indifferent,  and  the  answer  is 
[52 
[4  [13  |j_3  [13  [13' 

(3)  A  boat's  crew  consists  of  8  men,  of  whom  2  can  row 

only  on  the  stroke  side  of  the  boat,  and  3  can  row 
only  on  the  bow  side.  In  how  many  ways  can  the 
crew  be  arranged  ? 

There  are  left  3  men  who  can  row  on  either  side ;  2  of  the?3 
must  row  on  the  stroke  side,  and  1  on  the  bow  side. 

The  number  of  ways  in  which  these  three  can  be  divided  is 

_  =  3  ways. 

When  the  stroke  side  is  completed,  the  4  men  can  be  arranged 
in  [4  ways  ;  likewise,  the  4  men  of  the  bow  side  can  be  arranged 
in  [4  ways.  Hence  (Rule  II.)  the  arrangement  can  be  made  in 
3  X  [4  [4  =1728  ways. 

(4)  In  how  many  ways  can  10  copies  of  Homer,  6  of  Vir- 

gil, and  4  of  Horace  be  given  to  20  boys,  so  that  each 
boy  may  receive  a  book  ? 

The  boys  have  to  form  themselves  into  a  group  of  10  for  Ho- 
mer, of  6  for  Virgil,  of  4  for  Horace.     This  can  be  done  in 

[20 


[10  [6  [4 


different  ways. 


334  ALGEBRA. 


(5)  In  how  many  ways  can  3  copies  of  one  book,  2  of  an- 

other, and  1  of  a  third  be  given  to  a  class  of  12  boys, 
so  that  no  boy  shall  receive  more  than  1  book  ? 
In  every  possible  way  of  assigning  the  books,  6  boys  receive 
them.     These  6  may  be  selected  from  the  whole  12  (Rule  VIII.) 

in  1 12 

•  ■  ways. 
[6  [6       ^ 

When  thus  selected,  the  books  may  be  assigned  to  them  (Rule 
V.)  in  16 

Hence,  the  whole  number  of  ways  of  giving  the  books  is 
[12       _|6_ 

I 

(6)  In  how  many  ways  can  the  same  books  be  given  to  the 

12  boys,  so  that  no  boy  shall  receive  more  than  1 

copy  of  any  book  ? 

In  allotting  the  3  copies  of  the  first  book,  the  boys  are  to  be 
separated  into  two  groups  of  3  and  9 ;  and  the  group  of  3  will 
in  each  case  receive  the  books. 

|12 
This  can  be  done  in  -==~  ways. 
(3  [9       ^ 

Likewise  the  2  copies  of  the  second  book  may  be  given  in 

[12 

The  single  copy  of  the  third  book  can  be  given  in  12  ways. 
Hence  (Rule  II.)  the  books  may  be  given  in 
112         112 

(7)  In  how  many  ways  can  2  letters  be  selected  from  a,  h, 

c,  d,  e,/,  if  the  letters  may  be  repeated  in  making  the 

selection  ? 

Without  repetitions,  2  letters  can  be  selected  from  6  in  15  ways. 
With  repetitions,  as  aa,  etc.,  6  selections  can  be  made.  Hence, 
there  are  15  +  6  =  21  different  selections. 


CHOICE.  335 


(8)    In  how  many  ways  can  selections  of  3  letters  be  made 
from  a,  5,  c,  d,  e,/,  if  letters  may  be  repeated? 
These  selections  will  be  of  3  kinds : 
(i.)  All  three  letters  different. 
(ii.)  Two  letters  alike,  the  third  different, 
(iii.)  All  three  letters  alike. 
(By  Rule  VII.)  (i.)  gives  20  ways. 

(ii.)  gives  6  X  5  =  30 ;  for  we  can  choose  2  alike  in  6  ways, 
and  then  join  a  different  letter  to  each  pair  in  5  ways, 
(iii.)  gives  evidently  6  ways. 
Hence,  there  aro  in  all  20  +  30  +  6  =  56  different  selections. 

420.    This  may  also  be  shown  as  follows : 

By  adding  the  6  letters,  a,  h,  c,  d,  e,f,  to  each  selection  of 
the  kind  required,  they  will  become  selections  of  6  -f  3  =  9 
letters  out  of  6,  in  which  selections  each  letter  occurs  at 
least  once,  and  in  which,  therefore,  there  must  be  exactly  3 
repetitions.  These  repetitions  may  be  all  of  the  same  letter, 
or  divided  among  the  different  letters. 

Any  one  of  these  selections  may  be  made  by  writing  9 
places,  and  then  filling  them  in  alphabetical  order,  taking 
care  to  make  exactly  three  repetitions  in  passing  from  one 
end  of  the  row  to  the  other. 


Thus: 

1 

2 

3 

4 

5 

6 

1_ 

8 

a 

a 

a 

a 

h 

c 

"Y 

e 

^f 

•      a 

h 

c 

c 

c 

d 

e 

e 

f 

a 

h 

c 

c 

d 

d 

e 

e 

f 

By  striking  out  the  6  letters,  a,  b,  c,  d,  e,f,  each  once, 
there  is  left  in  the  first  row  aaa,  in  the  second  row  cce,  in  the 
third  row  cde ;  that  is,  three  of  the  required  selections. 

Now,  in  filling  the  9  places  for  each  row,  9  —  1  or  8  steps 
must  be  made,  of  which  exactly  3  are  repetitions,  and  each 
of  the  other  5  is  a  change  to  a  different  letter. 


336  ALGEBRA. 


The  3  repetitions  may  be  chosen  from  the  8  steps  (Rule 
VIII.)  in 

8x7x6      ^_ 

This,  then,  is  the  number  of  ways  in  which  3  letters  can 
be  selected  out  of  6  letters,  when  repetitions  are  allowed. 

In  other  words,  3  elements  can  be  selected  from  6  ele- 
ments, when  repetitions  are  allowed,  in  as  many  ways  as  3 
elements  can  be  selected  from  6  +  3  —  1,  or  8  elements 
without  repetitions. 

421.    Hence  the  general  rule  : 

XI,  The  number  of  ways  of  selecting  r  elements  from  n 
different  elements^  when  repetitions  are  allowed,  is  the  same 
as  the  number  of  ways  of  selecting  r  elements  from  n  +  r  —  1 
elements  without  repetitions. 

And  this  number  of  ways  is  (Rule  VIII.), 

|n  +  r-l      n{n+l){n-\-2) (n  +  r-1) 

|r  |n—  1    ~  [r 

(1)  In  how  many  ways  can  4  elements  be  selected  from  n 

elements,  when  repetitions  are  allowed  ? 

n(7i  +  l)(n  +  2)(n  +  3)     , 

li  ■         ' 

(2)  How  many  dominoes  are  there  in  a  set  numbered  from 

double  blank  to  double  nine  ? 

Each  domino  is  made  by  selecting  two  numbers  out  of  the  ten 
digits,  and  repetitions  are  included  ;  that  is,  the  two  numbers  on 
a  domino  may  be  the  same. 

Hence,  the  number  of  dominoes  is  equal  to  the  number  of  se- 
lections of  2  from  10  +  2  —  1,  or  11,  without  repetitions. 

^"^"  =  55.  An>. 
1X2 


CHOICE.  337 


(3)  In  how  many  ways  can  4  glasses  be  filled  with  5  kinds 

of  wine,  without  mixing  ? 

The  number  is  equal  to  the  number  of  ways  in  which  4  things 
can  be  selected  from  6  +  4  —  1,  or  8  things,  without  repetitions. 
5x6x7x8^^Q^^^ 
•  1x2x3x4 

(4)  In  how  many  ways  can  6  rugs  be  selected  at  a  shop 

where  2  kinds  of  rugs  are  sold  ? 

The  number  is  equal  to  the  number  of  ways  of  selecting  6 
from  2  +  6  —  1  =  7,  without  repetitions. 
2x3x4x5'x6x7 


1x2x3x4x5x6 
If  a  and  h  represent  the  two  kinds  of  rugs,  the  7  ways  are  as 


a  a  a  a  a  a  a  a  a  b  b  b 

a  a  a  a  a  b  a  a  b  b  b  b 

a  a  a  a  b  b  a  b  b  b  b  b 

bbbbbb 

422.    It  may  be  shown  that, 

XII.  The  number  of  ways  in  which  a  selection  (of  some,  or 
all)  can  be  made  from  n  different  things  is  2**  —  1. 

For  each  thing  can  be  either  taken  or  left,  that  is,  can  be  disposed 
of  in  two  ways. 

There  are  n  things  ;  hence  (Rule  II.)  they  can  all  be  disposed  of  in 
2»  ways.  But,  among  these  ways  is  included  the  case  in  which  all 
are  rejected  ;  and  this  case  is  inadmissable. 

Hence,  the  number  of  ways  of  making  a  selection  is  2**  —  1. 

(1)  In  a  shop  window  20  different  articles  are  exposed  for 

sale.     What  choice  has  a  purchaser  ? 
S^'- 1  =  1,048,575.  Ans. 

(2)  How  many  different  amounts  can  be  weighed  with  1  lb., 

2  lb.,  4  lb.,  8  lb.,  and  IG  lb.  weights? 

2^-1  =  31.  Am. 

(Let  the  student  write  out  the  31  weights.) 


338  ALGEBRA. 


423.    It  may  be  shown  that, 

XIII.    The  whole  number  of  ways  in  which  a  selection  can 

he  made  from  p  +  q^ +r things,  of  which  p  are  alike,  q  are 

alike,  r  are  alike,  etc.^  is  (p  +  1)  (c[+  1)  (r  + 1) —  1. 

For  the  set  of  p  things  may  be  disposed  of  in  p  +  1  ways,  since 
none  of  them  may  be  taken,  or  1,  2,  3, ,  ox  p,  may  be  taken. 

In  like  manner,  the  q  things  may  be  disposed  of  in  g'  +  1  ways ; 
the  r  things  in  r  +  1 ;  and  so  on. 

Hence  (Rule  II.)  all  the  things  may  be  disposed  of  in  {p  + 1)(2'  + 1) 
(r  + 1) ways. 

But  the  case  in  which  all  the  things  are  rejected  is  inadmissable ; 
hence,  the  whole  number  of  ways  is  (p  +  l)(g'  +  l)(r  +  l) —  1. 

(1)  In  how  many  ways  can  2  boys  divide  between  them  10 

oranges  all  alike,  15  apples  all  alike,  and  20  peaches 

all  alike  ? 

Here,  the  case  in  which  the  first  boy  takes  none,  and  the  case 
in  which  the  second  boy  takes  none,  must  be  rejected. 

Therefore,  the  answer  is  1  less  than  the  result,  according  to 
Rule  XIII.     11  X  16  X  21  -  2  =  3694.  Ans. 

(2)  If  there  be  m  kinds  of  things,  and  n  things  of  each  kind, 

in  how  many  ways  can  a  selection  be  made  ? 
In  this  case  p,  q,  r,  etc.,  are  all  equal,  and  each  is  equal  to  n. 
Hence,  the  result  is  (n  +  1)™  —  1. 

(3)  If  there  be  m  kinds  of  things,  and  1  thing  of  the  first 

kind,  2  of  the  second,  3  of  the  third,  and  so  on,  in 
how  many  ways  can  a  selection  be  made  ? 
|m  +  l  —  1.  Ans. 

Exercise  CXIX. 

1.  How  many  different  permutations  can  be  made  of  the 

letters  in  the  w^ord  Ecclesiastical,  taken  all  together  ? 

2.  Of  all  the  numbers  that  can  be  formed  with  four  of  the 

digits  5,  6,  7,  8,  9,  how  many  will  begin  with  56  ? 


CHOICE.  339 


3.  If  the  number  of  permutations  of  n  things,  taken  4  to- 

gether, be  equal  to  12  times  the  permutations  of  n 
things,  taken  2  together,  find  n. 

4.  With  3  consonants  and  2  vowels,  how  many  words  of  3 

letters  can  be  formed,  beginning  and  ending  with  a 
consonant,  and  having  a  vowel  for  the  middle  letter  ? 

5.  Out  of  20  men,  in  how  many  different  ways  can  4  be 

chosen  to  be  on  guard  ?  In  how  many  of  these  would 
one  particular  man  be  taken,  and  from  how  many 
would  he  be  left  out. 

6.  Of  12  books  of  the  same  size,  a  shelf  will  hold  5.     How 

many  different  arrangements  on  the  shelf  may  be 
made? 

7.  Of  8  men  forming  a  boat's  crew,  one  is  selected  as  stroke. 

How  many  arrangements  of  the  rest  are  possible? 
When  the  4  who  row  on  each  side  are  decided  on, 
how  many  arrangements  are  still  possible  ? 

8.  How  many  signals  may  be  made  with  6  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or  any 
number  at  a  time  ? 

9.  How  many  signals  may  be  made  with  8  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or  any 
number  at  a  time  one  above  another  ? 

10.  How  many  different  signals  can  be  made  with  10  flags, 

of  which  3  are  white,  2  red,  and  the  rest  blue,  always 
hoisted  all  together  and  one  above  another? 

11.  How  many  signals  can  be  made  with  7  flags,  of  which 

2  are  red,  1  white,  3  blue,  and  1  yellow,  always  dis- 
played all  together  and  one  above  another  ? 

12.  In  how  many  different  ways  may  the  8  men  serving  a 

field-gun  be  arranged,  so  that  the  same  man  may  al- 
ways lay  the  gun  ? 


340  ALGEBRA. 


13.  Find  the  number  of  signals  wliicli  can  be  made  with  4 

lights  of  different  colors  when  displayed  any  number 
at  a  time,  arranged  above  one  another,  side  by  side, 
or  diagonally. 

14.  From  10  soldiers  and  8  sailors,  how  many  different  par- 

ties of  3  soldiers  and  3  sailors  can  be  found  ? 

15.  IIow  many  signals  can  be  made  with  3  blue  and  2 

white  flags,  which  can  be  displayed  either  singly,  or 
any  number  at  a  time  one  above  another  ? 

16.  In  how  many  ways  can  a  party  of  6  take  their  places 

at  a  round  table? 

17.  Out  of  12  Democrats  and  16  Republicans,  how  many 

different  committees  can  be  formed,  each  consisting 
of  3  Democrats  and  4  Republicans  ? 

18.  From  12  soldiers  and  8  sailors,  how  many  different  par- 

ties of  3  soldiers  and  2  sailors  can  be  formed  ? 

19.  Find  the  number  of  combinations  of  100  things,  97  to- 

gether ? 

20.  With  20  consonants  and  5  vowels,  how  many  different 

words  can  be  formed  consisting  of  3  different  conso- 
nants and  2  different  vowels,  any  arrangement  of 
letters  being  considered  a  word  ? 

21.  Of  30  things,  how  many  must  be  taken  together  in  or- 

der that  having  that  number  for  selection,  there  may 
be  the  greatest  possible  variety  of  choice  ? 

22.  There  are  m  things  of  one  kind  and  n  of  another ;  how 

many  different  sets  can  be  made  containing  r  of  the 
first  and  s  of  the  second  ? 

23.  In  how  many  ways  may  10  persons  be  seated  at  a  round 

table,  so  that  in  no  two  of  the  arrangements  may  any 
one  have  the  same  neighbors  ? 


CHOICE.  341 


24.  The  number  of  combinations  of  n  things,  taken  r  to- 

gether, is  3  times  the  number  taken  r  —  1  together, 
and  half  the  number  taken  r-\-l  together.  Find  n 
and  r. 

25.  In  how  many  ways  may  12  things  be  divided  into  3 

sets  of  4  ? 

26.  How  many  words  of  6  letters  may  be  formed  of  3  vow- 

els and  3  consonants,  the  vowels  always  having  the 
even  places  ? 

27.  From  a  company  of  90  men,  20  are  detached  for  mount- 

ing guard  each  day.  How  long  will  it  be  before  the 
same  20  men  are  on  guard  together,  supposing  the 
men  to  be  changed  as  much  as  possible  ;  and  how 
many  times  will  each  man  have  been  on  guard  ? 

28.  Supposing  that  a  man  can  place  himself  in  3  distinct 

attitudes,  how  many  signals  can  be  made  by  4  men 
placed  side  by  side  ? 

29.  How  many  different  arrangements  may  be  made  of  11 

cricketers,  supposing  the  same  2  always  to  bowl  ? 

30.  Five  flags  of  different  colors  can  be  hoisted  either  singly, 

or  any  number  at  a  time  one  above  another.  How 
many  different  signals  can  be  made  with  them  ? 

31.  How  many  signals  can  be  made  with  5  lights  of  differ- 

ent colors,  which  can  be  displayed  either  singly,  or 
any  number  at  a  time  side  by  side,  or  one  above 
another  ? 

32.  The  number  of  permutations  of  n  things,  3  at  a  time,  is 

6  times  the  number  of  combinations,  4  at  a  time 
Find  n. 

33.  At  a  game  of  cards,  3  being  dealt  to  each  person,  any 

one  can  have  425  times  as  many  hands  as  there  are 
cards  in  the  pack.     How  many  cards  are  there  ? 


342  ALGEBRA. 


34.  In  how  many  ways  can  the  letters  of  the  word  logarithms 

be  arranged  : 
(i.)  Without  changing  the  order  of  the  vowels  ? 
(ii.)  Without  changing  the  place  of  any  vowels  ? 
(iii.)  Without  changing  the  relative  order  of  vowels 
and  consonants? 

35.  In  how  many  ways  can  the  letters  of  the  word  logarithms 

be  arranged  so  that  the  second,  fourth,  and  sixth  places 
may  be  occupied  by  consonants  ? 

•36.  In  how  many  ways  can  2  consonants  and  1  vowel  be 
chosen  out  of  the  letters  of  the  word  logarithms? 
And  in  how  many  of  these  will  the  letter  s  occur  ? 

37.  In  how  many  ways  can  a  set  of  12  black  and  12  white 

pieces  be  placed  on  the  black  squares  of  a  checker- 
board? 

38.  Of  the  numbers  from  10,000  to  100,000 : 

(i.)  In  how  many  is  every  digit  an  odd  number  ? 
(ii.)  In  how  many  is  every  digit  an  even  number  ? 
(iii.)  In  how  many  is  there  no  digit  lower  than  6  ? 
(iv.)  In  how  many  is  there  no  digit  higher  than  3  ? 
(v.)  How  many  contain  all  the  digits  1,  2,  3,  4,  5  ? 
(vi.)  How  many  contain  all  the  digits  0,  2,  4,  6,  8  ? 

39.  Out  of  20  consecutive  numbers  in  how  many  ways  can 

two  be  selected  whose  sum  shall  be  odd  ? 

40.  How  many  dominoes  are  there  in  a  set  numbered  from 

double  blank  to  double  twelve  ? 

41.  At  a  post-office  they  keep  10  kinds  of  postage-stamps. 

In  how  many  ways  can  a  person  : 
(i.)  Buy  twelve  stamps? 
(ii.)  Buy  eight  stamps  ? 
(iii.)  Buy  eight  different  stamps  ? 


CHAPTER    XXIII. 

Chance. 

424.  If  an  event  can  happen  in  a  ways  and  fail  in  h 
ways,  and  if  all  these  ways  are  equally  likely  to  happen  ;  if, 
also,  only  one  can  happen,  and  one  mtcst  happen,  then  the 
mathematical  probability  or  chance  of  the  event  happening 
is  expressed  by  the  fraction 


a  +  b 

I.  The  probability  of  an  event  happening  is  expressed  by 
the  fraction  whose  numerator  is  the  number  of  favorable 
ways,  and  denominator  the  whole  number  of  ways. 

Thus,  if  1  ball  be  drawn  from  a  bag  containing  3  white  balls  and 
9  black  balls,  the  chance  of  drawing  a  white  ball  is  -^^ ;  or,  as  it  is 
expressed,  one  chance  in  four. 

II.  The  probability  of  an  event  not  happening  is  expressed 
by  the  fraction  whose  numerator  is  the  number  of  unfavor- 
able ways,  and  denonfiinator  the  whole  number  of  ways. 

Thus,  if  a  denote  the  number  of  favorable  ways,  and  h  the  number 

of  unfavorable  ways,  then  the  fraction  will  express  the  proba- 

a  -\-h 
bility  of  the  event  not  happening.     If,  for  example,  1  ball  be  drawn 
from  a  bag  containing  3  white  and  9  black  balls,  the  chance  that  it 
will  not  be  a  white  ball  is  y%. 

425.    Since  _A.^  +  _^=:1, 

a-\-o      a-{-  0 

it  is  evident  that  the  chance  of  an  event  happening,  added 
to  the  chance  of  its  not  happening,  is  equal  to  1 ;  and,  since 


344  ALGEBRA. 


an  event  is  certain  to  happen  or  not  happen,  it  follows  that 
in  the  theory  of  chances 

III.  Certainty  is  expressed  hy  unity. 

426.  Since  -^  =  1 ~,  it  follows  that, 

a-{-o  a-{-o 

IV.  The  chance  of  an  event  not  happening  is  found  hy 
subtracting  from  unity  the  chance  that  it  does  happen. 

427.  If  the  number  of  favorable  ways  is  equal  to  the 
number  of  unfavorable  ways,  then 

The  chance  of  the  event 


ai-b      2a      ^ 
This  is  expressed  by  saying  "the  event  is  as  likely  to 
happen  as  not,"  or  "there  is  an  even  chance  for  the  event," 
or  "the  odds  are  even"  for  and  against  the  event. 

Thus,  in  tossing  a  cent,  there  is  an  even  chance  that  it  will  fall 
with  the  head  up. 

428.  If  a  >  b,  the  chance  of  the  event  happening  is  >  i. 
This- is  expressed  by  saying  "  the  event  is  probable,"  or 

"the  odds  are  as  a  to  ^  in  favor  of  the  event." 

If  a<.b,  the  chance  of  the  event  happening  is  <  i. 
This  is  expressed  by  saying  "the  event  is  improbable,"  or 

"the  odds  are  as  b  to  a  against  the  event." 

Thus,  the  odds  are  as  5  to  3  in  favor  of  drawing  a  white  ball  at  the 
first  trial  from  a  bag  containing  5  white  and  3  black  balls. 

Again,  since  a  die  has  6  faces,  on  one  of  which  is  an  ace,  the  chance 
for  an  ace  the  first  throw  is  ^ ;  and  the  odds  are  5  to  1  against  an  ace. 

429.  It  may  be  shown  that, 

Y.  If  there  are  several  events  of  ivhich  only  one  can  hap- 
pen, the  chance  that  some  one  of  them  will  happen  is  the 
sum  of  their  respective  chances  of  happening. 


CHANCE.  ol5 


For,  let  a,  b,  c denote  the  number  of  ways  favorable  to  the  first, 

second,  third event  respectively  ;  and  Ictp  denote  the  whole  num- 
ber of  ways,  all  equally  probable,  and  of  which  one,  and  only  one,  must 
happen.     Then  the  chances  of  the  first,  second,  third, events  are 

-,    — ,    -, respectively. 

P    P     P 

Since  there  are  a  +  b  +  c  + ways  favorable  to  some  one  or  other 

of  the  events  happening,  the  chance  in  favor  of  some  one  or  other  of 
the  events  is 

^  +  ^  +  ^  + or     ^+^  +  ^  + 

P  P      P      P 

If,  for  example,  a  bag  contain  3  white,  4  black,  5  red,  and  6  green 
balls,  the  chance  of  drawing  at  the  first  trial  a  white  or  a  black  ball 
^^  t\  +  t\  "=  tV  ;  ^^1®  chance  of  drawing  a  white  or  a  black  or  a  red 
ball  is  j\  +  tV  +  /?  ==  tI  ;  ^^^^  chance  of  drawing  a  white  or  a  black 
or  a  red  or  a  green  ball  is  y\  +  y\  +  y\  +  ^^y  =  ||  =  1 ;  that  is,  certainty. 

(1)  When  two  dice  are  thrown,  what  is  the  chance  of  throw- 

ing double  aces? 

Each  die  may  fall  in  any  one  of  6  ways ;  therefore  both  dice 
in  6  X  6  =  36  ways  (^  403).  Of  these  ways  only  one  will  give 
double  aces.     Hence,  the  chance  of  double  aces  =  ■^.   Ans. 

(2)  What  is  the  chance  of  throwing  doublets  in  a  single 

throw  with  two  dice  ? 

The  dice  may  fall  in  36  ways.     Of  these,  6  will  be  doublets. 

Hence,  the  chance  of  throwing  doublets  =  /g-  =  ^.  Ans. 

(3)  What  is  the  chance  of  throwing  a  six  and  a  five  by  a 

single  throw  of  two  dice  ? 

The  dice  may  fall  in  36  ways.  Of  these  ways  the  first  die  may 
turn  up  a  six  and  the  second  a  five,  or  the  first  may  turn  up  a 
five  and  the  second  a  six.     Hence,  the  chance  is  ^\  =  -j^j.  Ans. 

(4)  With  two  dice,  what  is  the  chance  of  making  a  throw 

so  that  one  and  only  one  die  may  turn  up  o^five  ? 

In  6  of  the  36  possible  ways  one  die  will  turn  up  a.  five,  and  the  ' 
other  also  will  turn  up  a  five  in  6  ways.     One  of  these  12  ways 
will  be  double  fives;  so  that  there  are  11  ways  in  which  one  die, 
and  only  one,  will  turn  up  a.  five,  and  the  chance  is  \\.  Ans. 


346  ALGEBRA. 


(5)  What  is  the  chance  of  making  a  throw  that  will  amount 

to  five  f 

Of  the  36  possible  ways,  1  and  4,  4  and  1,  2  and  3,  3  and  2 
amount  to  jive.     Hence,  the  chance  is  -^  =  \.  Ans. 

(6)  In  a  single  throw  with  two  dice,  if  the  player  may 

count  the  number  on    one  of  the  dice,  or  the  sum 
of  the  numbers  on  the  two  dice,  what  is  the  chance 
of  throwing  yi-ye  / 
The  chance  is  |^  +  ^  =  |f .  Ans. 

(7)  If  A's  chance  of  winning  a  prize  is  ^,  and  B's  ^,  what 

is  the  chance  that  neither  will  obtain  a  prize  ? 
The  chance  that  one  will  win  is  ^  +  J  =  ■^■^.    Hence,  the  chance 
that  neither  will  win  is  1  —  -^j  =  ||.  Ans. 

(8)  If  4  cards  are  drawn  from  a  pack  of  52  cards,  what  is 

the  chance  that  there  will  be  one  of  each  suit  ? 
Four  cards  can  be  selected  (§  415)  from  the  pack  in 
52X51X50X49^  , 

1X2X3X4  ^ 

But  4  cards  can  be  selected  so  as  to  be  one  of  each  suit  in 
134  =  28,561  ways.  (g  404.) 

Hence,  the  chance  is  ^fyWVs"  "==  tV-  nearly. 

(9)  If  4  cards  are  drawn  from  a  pack,  what  is  the  chance 

that  they  will  be  the  4  aces  ? 

There  are  [4  =  24  ways  of  drawing  the  four  aces,  and  270,725 
ways  of  drawing  four  cards.  Hence,  the  chance  is  ^yf f^-j,  or  1 
chance  in  11,280. 

(10)  Three  balls  are  to  be  drawn  from  an  urn  containing  5 

black,  3  red,  and  2  white  balls.     What  is  the  chance 

of  drawing  1  red  and  2  black  balls  ? 

10  X  9  X  8 

Three  balls  can  be  selected  from  the  whole  10  in  — — - 

1x2x3 

==  120  ways.    Also,  2  black  balls  can  be  selected  from  the  5  black 

balls  in  — ^ —  =  10  ways,  and  1  red  ball  from  the  3  red  balls  in  3 
1X2  *" 


CHANCE.  347 


ways.  Hence,  1  red  and  2  black  balls  can  be  drawn  in  3  X 10  = 
30  ways.  That  is,  there  are  120  different  ways  of  drawing  3 
balls,  and  30  of  these  ways  give  1  red  and  2  black  balls. 

The  chance,  then,  of  1  red  and  2  black  balls  is  ^^j^  =  ^.   Ans. 

(11)  If  2  tickets  are  drawn  from  a  package  of  30  tickets, 

marked  1,  2,  3, ,  what  is  the  chance  that  both  will 

be  marked  with  odd  numbers  ? 

30  X  29 
Two  tickets  can  be  drawn  from  30  tickets  in  — — —  ways ; 

1x2        \' 
and  2  odd  numbers  can  be  drawn  from  the  15  odd  numbers  in 

15  X  14  .   15  X  14 

— - —  ways.     Hence,  the  chance  is =  J^.  Ans. 

1X2        -^  '  30x29     ^ 

(12)  In  a  bag  are  5  white  and  4  black  balls.     If  they  are 

drawn  out  one  by  one,  what  is  the  chance  that  the 
first  will  be  white,  the  second  black,  and  so  on,  al- 
ternately ? 

The  9  balls  can  be  arranged  in  [9  ways.  The  5  white  balls 
can  be  arranged  in  the  odd  places,  and  the  4  black  balls  in  the 
even  places,  in  |_5  x  [4  ways.     Hence,  the  chance  of  alternate 

0'^"^'^  15x14 

<13)  From  a  bag  containing  10  balls,  4  are  drawn  and  re- 
placed, then  6  are  drawn.  Find  the  chance  that 
the  4  first  drawn  are  among  the  6  last  drawn. 

The  second  drawing  could  be  made  altogether  in 

UP 
^  =  210  ways. 

But  the  drawing  can  be  made  so  as  to  include  the  4  first  drawn 

in  16 

_=  15  ways. 

since  the  only  choice  consists  in  selecting  2  balls  from  the  6  not 
previously  drawn.     Hence,  the  chance  is  -^^^  =  ^.  Ans. 


348  ALGEBRA. 


(14)  The  chance  of  an  event  is  ^.     What  are  the  odds  in 

favor  of  the  event  ? 

4  to  3.  Ans. 

(15)  The  odds  against  an  event  are  3  to  1.     What  is  the 

chance  of  the  event? 
1-.  Ans. 

(16)  The  odds  against  an  event  are  m  to  n.     What  is  the 

chance  of  the  event  ? 


m  +  n 


(17)  If  4  coppers  are  tossed,  what  is  the  chance  that  exactly 

2  will  turn  up  heads  ? 

Since  each  coin  may  fall  in  2  ways,  the  4  coins  may  fall  in 
2*=  16  ways.    The  2  coins  to  turn  up  heads  can  be  selected  from 

4x3 
the  4  coins  in =  6  ways.     Hence,  the  chance  is  xV  "=  f  5 

1  X  ^ 
and  the  odds  are  5  to  3  against  it. 

(18)  A  has  3  tickets  in  a  lottery  where  there  are  3  prizes 

and  6  blanks.    Find  his  chance  of  winning  one  prize, 
two  prizes,  three  prizes,  respectively. 

9x8x7 
Three  tickets  can  be  selected  from  9  tickets  in = 

1x2x3 
84  ways.     A  prize  ticket  can  be  selected  from  the  3  prize  tick- 
ets in  3  ways,  and  2  blanks  can  be  selected  from  the  6  blanks 

in  — — —  =  15  ways ;  therefore,  1  prize  and  2  blank  tickets  can 

be  selected  in  3  X  15  =  45  ways.     Hence,  the  chance  of  draw- 
ing one  prize  is  ff . 

Again,  1  blank  and  2  prize  tickets  can  be  selected  in  6  X 

3x2 

— ^ —  =  18  ways.     Hence,  the  chance  of  two  prizes  is  ^f. 

J-  X  ^ 

Also,  the  3  prize  tickets  can  be  selected  in   only  1  way. 

Hence,  the  chance  of  drawing  three  prizes  is  g^j. 


CHANCE.  349 


(10)    What  is  the  chance  that  A  in  Ex.  18  wins  at  least  one 
prize  ? 

The  chance  is  |^  +  if  +  ^^  =  f  f  =  if-  ^^r,  he  will  have  at 
least  one  prize  in  any  one  of  the  three  cases  given  in  (18). 

Or,  the  chance  may  be  found  in  this  way :  A  gets  a  prize  un- 
less his  three  tickets  all  turn  out  blanks.  Three  tickets  can  be 
selected  from  the  whole  number  in  84  ways,  and  from  the  6 

6x5x4 
blanks  in  —     ^     —  =  20  ways.     Hence,  the  chance  that  they 

i  X  ^  X  C) 
will  all  be  blank  is  ||  =  ^j ;  and  the  chance  against  this  result 

isl-^V=l4. 


430.  If  a  person  is  to  receive  a  prize  in  case  a  particular 
event  happens,  the  sum  of  money  for  which  he  may  equit- 
ably sell  his  chance  for  the  prize  is  called  his  expectation 
from  the  event. 

YI.  The  expectation  from  an  uncertain  event  is  the  product 
of  the  chance  that  the  event  will  happen  hy  the  sum  to  he 
realized  in  case  the  event  happens. 

Thus,  if  there  is  a  lottery  with  40  tickets,  and  1  prize  worth  $  100, 
a  person  might  equitably  pay  §  100  for  the  whole  40  tickets,  since  one 
of  them  is  sure  to  draw  the  $  100.  Now,  all  of  the  tickets  are  of  equal 
value  before  the  drawing ;  hence  the  value  of  each  ticket  is  -^  of 
$100.  The  value  of  5  tickets  is  j\  of  |100,  or  $12.50;  that  is,  the 
product  of  the  chance  which  the  holder  of  5  tickets  has  of  winning 
the  prize  and  the  value  of  the  prize ;  that  is,  ^^  of  $  100. 

(20)  If  a  lottery  has  1  prize  of  $  50,  2  prizes  of  $  10  each, 
4  prizes  of  $  1  each,  and  13  blanks,  what  is  the  ex- 
pectation of  the  holder  of  1  ticket  ? 

The  chance  of  drawing  the  prize  of  $50  is  ^j^,  and  the  expecta- 
tion is  -^V  of  1 50  =  $  2.50.  The  chance  of  drawing  a  prize  of 
$5  is  ^2^,  and  the  expectation  is  ^  of  $5  =  $.50.  The  chance 
of  drawing  a  prize  of  $1  is  /^,  and  the  expectation  is  /^  of  $1 
=  $  .20.  Hence,  the  whole  expectation  is  $  2.50  +  $  .50  +  $  .20  = 
$3.20.  Ans. 


350  ALGEBRA. 


Exercise  CXX. 

1.  If  I  throw  a  single  die,  what  is  the  chance  that  it  will 

turn  up  : 
(i.)   An  ace  ? 
(ii.)    An  ace  or  a  two  ? 
(iii.)    Neither  an  ace  nor  a  two  ? 

2.  The  chance  of  a  plan  succeeding  is  i.     What  is  the 

chance  that  it  fails  ? 

3.  If  the  odds  are  10  to  1  against  an  event,  what  is  the 

probability  of  its  happening  ? 

4.  If  the  odds  are  5  to  2  in  favor  of  the  success  of  an  ex- 

periment, what  are  the  respective  chances  of  success 
or  failure  ? 

5.  The  chance  of  an  event  is  f.     Find  the  odds  for  or 

against  the  event. 

6.  What  is  the  chance  of  a  year,  not  a  leap-year,  having 

53  Sundays? 

7.  Two  numbers  are  chosen  at  random.     Find  the  chance 

that  their  sum  is  even. 

8.  If  4  cards  are  drawn  from  a  pack,  what  is  the  chance 

that  they  will  all  be  hearts  ? 

9.  If  10  persons  stand  in  a  line,  what  is  the  chance  that 

2  assigned  persons  will  stand  together  ? 

10.  If  10  persons  form  a  ring,  what  is  the  chance  that  2 

assigned  persons  will  stand  together  ? 

11.  Show  that,  if  n  persons  sit  down  at  a  round  table,  the 

odds  against  2  particular  persons  sitting  next  to  each 
other  are  w  —  3  to  2. 


CHANCE.  351 


12.  If  2  letters  are  selected  at  random  out  of  the  alphabet, 

what  is  the  chance  that  both  will  be  vowels  ? 

13.  Five  men,  A,  B,  C,  D,  E,  speak  at  a  meeting,  and  it  is 

known  that  A  speaks  before  B.  What  is  the  chance 
that  A  speaks  immediately  before  B  ? 

14.  A,  B,  0  have  equal  claims  for  a  prize.     A  says  to  B, 

"  You  and  I  will  draw  lots,  and  the  winner  shall  draw 
lots  with  0  for  the  prize."    Is  this  fair? 

15.  A  person  is  allowed  to  draw  2  tickets  from  a  bag  con- 

taining 40  blank  tickets,  and  10  tickets  each  entitling 
the  holder  to  a  prize  of  §  100.  What  is  his  expecta- 
tion ? 

16.  One  of  two  events  must  happen.     If  the  chance  of  one 

is  I  of  that  of  the  other,  find  the  odds  on  the  first. 

17.  There  are  3  events.  A,  B,  C,  one  of  which  must  happen. 

The  odds  are  3  to  8  on  A,  and  2  to  5  on  B.  Find 
the  odds  on  C. 

18.  In  a  bag  are  7  white  and  5  red  balls.     Find  the  chance 

that  if  one  is  drawn  it  will  be  (i.)  white  or  (ii.)  red ; 
or,  if  two  are  drawn,  that  they  will  be  (i.)  both  white, 
(ii.)  both  red,  or  (iii.)  one  white  and  the  other  red. 

19.  If  3  cards  are  drawn  from  a  pack,  what  is  the  chance 

that  they  will  be  king,  queen,  and  knave  of  the  same 
suit  ? 

20.  A  general  orders  2  men  by  lot  out  of  100  mutineers  to 

be  shot;  the  real  leaders  of  the  mutiny  being  10 
in  number.  Find  the  chance  (i.)  that  one,  (ii.)  that 
two  of  the  leaders  will  be  shot. 

21.  Show  that  the  odds  are  8  to  1  against  throwing  9  in  a 

single  throw  v.'iih  2  dice. 


852  ALGEBRA. 


22.  Show  that  in  a  throw  with  3  dice  the  chance  of  either  a 

triplet  or  a  doublet  is  ^. 

23.  In  a  bag  are  5  white  and  4  black  balls.     If  drawn  out, 

one  by  one,  what  is  the  chance  that  the  first  will  be 
white,  the  second  black,  and  so  on,  alternately  ? 

24.  A  bag  contains  2  white  balls,  3  black  balls,  and  5  red 

balls.     If  4  balls  are  drawn,  find  the  chance  that 
there  shall  be  among  them  : 
(i.)   Both  the  white  balls, 
(ii.)    Two  onl^/  of  the  black  balls, 
(iii.)    Two  at  least  of  the  red  balls. 

431.  A  series  of  events,  such  that  ouIt/  one  of  them  can 
happen,  may  be  called  a  series  of  exclusive,  or  dependent, 
events. 

Two  or  more  events,  such  that  both  or  all  may  happen, 
are  called  non-exclusive,  or  independent,  events. 

Thus,  if  a  copper  be  thrown  twice  in  succession  it  may  fall  head 
up  both  times ;  and,  if  it  be  thrown  ten  times,  it  is  possible  for  it  to 
fall  head  up  each  time. 

432.  If  there  are  two  or  more  independent  events,  the. 
occurrence  of  all  of  them  simultaneously  or  in  succession 
may  be  regarded  as  a  single  compound  event. 

Thus,  in  tossing  a  copper  twice,  the  event  of  its  falling  with  head 
up  at  both  trials  may  be  regarded  as  an  event  compounded  of  two 
simple  events ;  namely,  with  head  up  at  the  first  trial,  and  with  head 
up  at  the  second  trial. 

(1)    In  tossing  a  copper  twice,  what  is  the  chance  of  its 

falling  head  up  both  times  ? 

The  chance  of  a  head  at  each  trial  is  J.  If  these  separate 
chances  were  added  (according  to  Rule  V.),  the  result  would  be 
1 ;  that  is,  certainty  ;  a  result  obviously  false.  Rule  V.  applies 
only  to  dependent  or  exclusive  events.  In  this  case,  however, 
the  events  are  independent,  or  non-exclusive. 


CHANCE.  353 


Now,  each  time  the  copper  is  thrown,  it  can  fall  in  2  ways. 
Hence,  the  double  fall  can  occur  in  2  X  2  =  4  ways :  |  403. 

1.  Both  times  a  head. 

2.  First  time  a  head,  second  time  a  tail, 

3.  First  time  a  tail,  second  time  a  head. 

4.  Both  times  a  tail. 

Only  one  of  these  four  ways  gives  heads  both  times.  Hence, 
the  chance  of  heads  both  times  is  ^  =  J  X  J ;  that  is,  the  product 
of  the  separate  chances  of  a  head  at  each  trial. 

433.  In  general, 

VII.  The  chance  that  two  independent  events  both  happen 
is  the  product  of  their  separate  chances  of  happening. 

For  the  product  of  the  denominators  of  the  separate  chances  is  the 
whole  number  of  ways  in  which  the  compound  event  can  happen ; 
and  the  product  of  the  numerators  is  the  number  of  ways  favorable 
to  its  happening. 

(2)  A  bag  contains  3  balls,  two  of  which  are  white  ;  another 
contains  6  balls,  five  of  which  are  white.  If  a  person 
draws  1  ball  from  each  bag,  what  is  the  chance  that 
both  balls  drawn  will  be  white  ? 

The  first  ball  can  be  drawn  in  3  ways  and  the  second  in  6 
ways.  Hence,  both  can  be  drawn  in  3  X  6  =  18  ways.  Also, 
the  first  ball  can  be  a  white  ball  in  2  ways,  and  the  second  in  5 
ways.  Hence,  they  can  be  both  white  in  2  X  5  =  10  ways.  The 
chance  of  both  white  therefore  is  -J^|  =  f  X  f  ;  that  is,  the  product 
of  the  separate  chances  of  a  white  ball  at  each  trial. 

434.  In  like  manner,  whatever  the  number  of  simple 
events  that  unite  to  produce  a  compound  event,  it  may  be 
shown  that : 

VIII.  The  chance  of  a  compound  event  is  the  product  of 
the  separate  chances  of  the  simple  events  that  unite  to  produce 

it 

Note.  It  is  important  not  to  confound  exclusive  events  with  non- 
exclusive, and  not  to  apply  Rule  V.  to  problems  to  which  Rule  VII. 
applies. 


354  ALGEBEA. 


(3)  The  chance  that  A  can  solve  a  given  problem  is  |-,  and 

the  chance  that  B  can  solve  it  is  -f^;-    If  both  try,  what 

are  the  chances  (i.)  that  both  solve  it;  (ii.)  that  A 

solves  it  and  B  fails;  (iii.)  that  A  fails  and  B  solves 

it ;  (iv.)  that  both  fail  ? 

A's  chance  of  success  is  f ,  A's  chance  of  failure  is  ^. 

B's  chance  of  success  is  j^,  B's  chance  of  failure  is  -^^^ 

Therefore,  the  chance  of    (i.)  is  f  X  x\  =  ^ 

the  chance  of  (ii.)  is  f  x  jV  =  \ 

the  chance  of  (iii.)  is  J  X  -3^2  =  i 

the  chance  of  (iv.)  is  J  x  -j^^  =  iV 
The  sum  of  these  four  chances  is  |f  +  -jl  +  /^  +  -jV  =  1,  as  it 
ought  to  be,  since  1  of  the  4  results  is  certain  to  happen. 

(4)  In  Ex.  (3)  what  is  the  chance  that  the  problem  will  be 

solved  ? 

The  chance  that  both  fail  is  ■^.  Hence,  the  chance  that  both 
do  not  fail,  or  that  the  problem  will  be  solved,  is  1  —  -^^  =  ff. 

(5)  There  are  3  bags,  the  first  containing  1  white  and  1 

black  ball ;  the  second,  1  red  and  2  white  balls ;  the 
third,  3  white  and  2  green  balls.     If  a  person  draw 
a  ball  from  each  bag,  what  is  the  chance  that  all 
three  balls  drawn  will  be  white  ? 
i  X  f  X  I  =  i.  Ans. 

(6)  Under  the  conditions  of  the  last  problem,  what  is  the 

chance  that  no  one  of  the  balls  drawn  will  be  white  ? 

The  chances  of  failing  to  draw  a  white  ball  at  the  three  trials 
are  ^,  J,  |,  respectively.  Therefore,  the  chance  of  faiUng  alto- 
gether is  J  X  i^  X  f  =  yV-'  ^^s- 

(7)  What  is  the  chance  in  Ex.  (5)  of  drawing  at  least  one 

white  ball  ? 

One  white  ball  will  be  drawn  unless  all  three  trials  fail.  The 
chance  that  all  three  fail  is  ■^^.  Therefore,  the  chance  of  draw- 
ing at  least  one  white  ball  is  1  —  yV  =  tI    Ans. 


CHANCE.  355 


(8)    What  is  the  chance  in  Ex.  (5)  that  one,  and  only  one, 

white  ball  should  be  drawn  in  the  three  trials  ? 

The  chance  of  a  white  ball  from  the  first  bag  and  not 
from  the  others  is i  X  ^  X  f  =  g^j 

The  chance  of  a  white  ball  from  the  second  bag- and  not 
from  the  others  is J  X  f  X  f  =  /^r 

The  chance  of  a  white  ball  from  the  third  bag  and  not 
from  the  others  is i  X  ^  X  f  =  TfV 

Therefore,  the  sum  of  these  chances  is /^ 


(9)  When  6  coins  are  tossed,  what  is  the  chance  that  at 

least  one  will  fall  with  the  head  up  ? 
The  chance  that  all  will  fall  heads  down  is^xjxjxjxjx^ 
=  -^^.    Hence,  the  chance  that  this  will  not  happen  is  1  —  ^j  =  ||. 

(10)  When  6  coins  are  tossed,  what  is  the  chance  that  one, 

and  only  one,  will  fall  with  the  head  up  ? 

The  chance  that  the  first  alone  falls  with  head  up  is  J  x  J  X  i^ 
XjxJxJ  =  ^i;  the  chance  that  the  second  alone  falls  with 
the  head  up  is  -^^  ;  and  so  on. 

Hence,  the  chance  that  some  one,  and  only  one,  falls  head  up 
is  z\  +  ^i  +  ^V  +  ^?  +  uV  +  zh  ==  /i  =  A-  ^ns. 

(11)  When  4  dice  are  thrown,  what  is  the  chance  that  two, 

and  only  two,  turn  up  aces  ? 

The  chance  that  any  particular  two  of  the  4  dice  turn  up  aces, 
and  the  other  two  something  else,  isixix|^Xf  =  tI  f^- 

Now  the  number  of  ways  in  which  this  can  happen  is  the 
number  of  ways  in  which  2  dice  can  be  selected  from  4  dice, 
or  6  ways.  Hence,  the  chance  that  two,  and  only  two,  turn  up 
aces  is  6  X  yf  ^^  =  ^W-  Ans. 

(12)  When  4  dice  are  thrown,  what  is  the  chance  that  they 

will  all  turn  up  alike  ? 

The  chance  that  the  first  and  second  turn  up  alike  is  \. 

The  chance  that  the  third  turn  up  like  the  first  and  second  is  \. 

The  chance  that  the  fourth  turn  up  like  the  others  is  J. 

Hence,  the  chance  that  the  four  turn  up  alike  is  ^|-^.  Ans. 


356  ALGEBRA. 

(13)  When  4  dice  are  thrown,  what  is  the  chance  that  two, 

and  only  two  of  them,  should  turn  up  alike  ? 

The  chance  that  any  two  should  be  alike  is  I,  the  chance 
that  the  third  should  be  different  is  |,  and  the  chance  that  the 
fourth  should  be  different  from  all  the  rest  is  |.  Hence,  the 
chance  that  a  pair  should  agree  while  the  others  should  differ 
from  the  pair  and  from  each  other  is  ^  X  |  X  f  =  jjs-     The  pair 

4x3 
to  agree  may  be  selected  in  =  6  ways.     Hence,  the  total 

chance  6  X  2tV  ==  f-  '^'^^^ 

(14)  When  4  dice  are  thrown,  what  is  the  chance  that  they 

should  all  fall  different  ? 

The  chance  that  the  second  should  differ  from  the  first  is  f , 
that  the  third  should  differ  from  both  the  first  and  the  second  is 
f ,  and  that  the  fourth  should  differ  from  all  the  others  is  |. 
Hence,  the  required  chance  is  f  X  f  X  f  =  j\.  Ans. 

(15)  A  single  die  is  thrown  until  it  turns  up  an  ace.    What 

is  the  chance  that  it  must   be  thrown  at  least  10 

times  ?    What  is  the  chance  that  it  must  be  thrown 

exactly  10  times? 

The  chance  of  failing  the  first  9  times  is  (Rules  IV.  and  VII.) 
(I)'.  This,  then,  is  the  chance  that  at  least  10  trials  must  be 
made.  Since  (|)^  is  the  chance  of  failing  the  first  9  trials,  and 
\  the  chance  of  success  the  next  trial ;  therefore  (Rule  VII.)  (f  )* 
X  i  is  the  chance  that  exactly  10  throws  must  be  made 

(16)  What  is  the  chance  that  a  person  with  2  dice  will 

throw  double  aces  exactly  3  times  in  5  trials  ? 

The  chance  of  throwing  double  aces  at  any  particular  trial  is 
i  X  i  =^  T?'  ^°^  <^f  failing  is  ||.  Hence,  the  chance  of  succeeding 
at  3  assigned  trials,  and  failing  at  the  other  2  trials,  is  i^-^)^  X  (|f  f. 
Now  double  aces  will  be  thrown  exactly  three  times  if  thrown  in 
any  set  of  3  trials  that  may  be  assigned  out  of  the  5  trials,  and  fail 
in  the  other  2  trials.     3  trials  can  be  assigned  out  of  5  trials 

in  ^^^^  ^  ==  10  ways.    Hence,  the  chance  is  {-^^f  x  (||f  X 10. 
1  X  .^  X  3 


CHANCE.  357 


(17)  A  and  B  throw  with  a  single  die  alternately,  A  throw- 

ing first ;  and  the  one  who  throws  an  ace  first  is  to 

receive  a  prize  of  $  10.     What  are  their  respective 

expectations  ? 

The  chance  for  the  prize  at  the  first  throw  is  ^  ;  at  the  second, 
I X  i;  at  the  third,  (|f  X  i ;  at  the  fourth,  (l)^  x  ^ ;  and  so  on. 

As  A  has  the  first,  third,  etc.,  and  B  the  second,  fourth,  etc., 
throws, 

^'»  ^^^^"^^  =     ^+(f)'Qn  + _  6  .  so  that 

B's  chance     |  of  i  +  (f  f  of  ^ -i- ^ 

A's  expectation  is  j\  of  $10  =  ?5^j,  and 
B's  expectation  is  y\  of  $10  =  $4/x- 

(18)  A  and  B  play  at  a  game  that  cannot  be  a  drawn 

game,  and  on  an  average  A*  wins  3  games  out  of  5 

games.     Out  of  5  games,  what  is  the  chance  that  A 

wins  at  least  three  ? 

The  chance  that  A  wins  3  assigned  games  out  of  5  games  is 
(1/  X  (5)*  ==  AVj-  The  3  games  may  be  assigned  in  10  ways. 
Hence,  A's  chance  for  3  games  is  10  X  -^VW  =  Mf  §• 

The  chance  that  A  wins  4  games,  and  B  the  other  game,  is 
■ii^  X  5  =  /xsV-     The  chance  that  A  wins  all  the  games  is  (f  )* 

For  A  to  win  at  least  3  games,  he  must  win  3,  4,  or  5  games. 
Hence,  A's  chance  for  at  least  3  games  is  |f  |§  +  /j^**^-  +  -^^^^ 

(19)  A's  skill  at  a  game,  which  cannot  be  a  drawn  game, 

is  to  B's  skill  as  3  to  4.  If  they  play  3  games,  what 
is  the  chance  that  A  will  win  more  games  than  B  ? 
Their  respective  chances  of  winning  a  particular  game  are 
^  and  f .  For  A  to  win  more  games  than  B,  he  must  win  all  3 
games  or  2  games.  The  chance  that  A  wins  all  three  is  (f )'  = 
■jY^--  The  chance  that  A  wins  any  assigned  set  of  2  games  out 
of  the  3  games,  and  that  B  wins  the  other,  is  {^f  x  ^.  As  there 
are  3  ways  of  assigning  a  set  of  2  games  out  of  3,  the  chance 
that  A  wins  2  games,  and  B  the  other,  is  {^f  x  f  X  3  =  ^f f . 
Hence,  the  chance  that  A  wins  more  than  B  is  /j'j  +  ||f  =  ||f . 


358  ALGEBRA. 


(20)  In  the  last  example,  find  B's  chance  of  winning  more 

games  than  A. 

B's  chance  of  winning  all  three  games  is  (|)^  =  -^^-g.  His 
chance  of  winning  2  games,  and  A  the  other  game,  is  (i)^  x  f 
X  3  =  l^lf .     Hence,  his  chance  of  winning  more  games  than  A 

IH     64      I     144  _  208 
^^  JiJ  +  3  ?3  —  3¥3- 

Notice  that  A's  chance  added  to  B's  chance,  i||  +  |ff  =1. 
Why  should  this  be  so  ? 

(21)  A  plays  a  set  of  games  (drawn  games  excluded)  with 

B,  his  chance  of  winning  a  single  game  being  to  B's 
as  3  :  2.     "What  is  the  probability : 
(i.)  That  A  will  win  4  games  at  least  out  of  7  ? 
(ii.)  That  he  will  win  4  games  before  B  wins  3  ? 

(i.)  A's  chance  of  winning  a  single  game  is  |,  and  B's  chance 
is  ^.  The  chance  that  A  wins  at  least  4  games  out  of  6  is  the 
sum  of  the  chances  that  he  wins  4,  5,  G,  or  7  games  out  of  7,  in 
any  possible  order. 

The  chance  that  he  wins  all  7  games  =  (f f. 

The  chance  that  he  wins  6  games       =  7  (|f  X  f . 

The  chance  that  he  wins  5  games       =^  21  (f  f  x  (|)*. 

The  chance  that  he  wins  4  games       =  35  (f)''  x  (f)^. 

The  sum  of  these  values  =  — — ^.  Ans. 

5^ 

(ii.)  Here  the  chance  required  is  that  A  shall  win  at  least  4 

games ;  that  is,  4,  5,  or  6  games  out  of  6. 

The  chance  that  he  wins  all  6  games  =  (|)^, 

The  chance  that  he  wins  5  games        =  6  X  (f  )^  X  |. 

The  chance  that  he  wins  4  games        =  15  x  (f )*  X  (f)^. 

The  sum  of  these  values  = .  Ans. 

5^ 

(22)  In  a  certain  locality  it  is  found  that,  on  the  average 

for  10  years,  out  of  100  persons  40  years  old  at  the 
beginning  of  the  decade,  20  die ;  out  of  100  persons 
50  years  old,  30  die ;  and  out  of  100  persons  60 
years  old,  40  die.  What  is  the  odds  against  a  per- 
son 40  years  old  living  30  years  longer  ? 


CHANCE.  359 


The  chance  that  he  dies  between  40  and  50  is  ^ ;  that  he  lives 
till  50,  and  dies  between  50  and  60,  is  f  X  j^  =  /j ;  that  he  lives 
till  60,  and  dies  between  60  and  70,  is  |-  X  xV  X  j^  =  t A-  Hence, 
the  chance  that  he  dies  between  40  and  70  is  ^^  +  j^  +  xV^  = 
■^j'y.  Therefore,  the  odds  against  his  living  for  30  years  are  83 
to  42,  or  about  2  to  1. 

(23)    A  is  40  years  old  and  B  50  years  old.     "What  is  the 
probability  that  at  least  one  of  them  will  be  alive  10 
years  hence  ? 
The  chance  that  A  dies  is  ^,  and  the  chance  that  B  dies  is  ^jj. 

Hence,  the  chance  that  both  die  is  ^  X  jjj  =  3%  ;  and  the  chance 

that  one  at  least  will  be  alive  is  1  —  -^j^  =  ^|. 

434.  Cases  often  occur  where  the  simple  events  which 
unite  to  form  the  compound  event  are  so  related  that  the 
happening  of  one  of  them  alters  the  chances  of  the  others. 

(1)  What  is  the  chance  of  drawing  in  succession  2  vowels 

from  the  alphabet  ? 

The  chance  of  drawing  a  vowel  the  first  time  is  -^^ ;  but,  if 
one  vowel  is  drawn,  the  chance  of  drawing  another  is  1^^.  Hence, 
(Rule  IV.)  the  required  chance  is  /^  X  ^  =  ^. 

(2)  A  bag  contains  5  white  and  6  black  balls.     What  is 

the  chance  of  drawing  5  times  in  succession  a  white 
ball,  the  balls  drawn  not  being  replaced  ? 
Tr  X  tV  X  f  X  f  X  f  =  5^.  Ans. 

(3)  What  would  have  been  the  chance  in  the  last  example 

if  after  each  drawing  the  ball  had  been  replaced  ? 
{j\f.  Ans. 

(4)  If  the  chance  of  an  event  at  first  is  as  a  to  b,  and  if 

whenever  it  happens,  the  number  of  favorable  ways, 
as  well  as  the  whole  number  of  ways,  is  diminished 
by  unity ;  find  the  chance  that  the  event  will  occur  n 
times  in  succession. 


a(^-l)(a-2) (g-n  +  l)     . 

b{b-l){b-2) {b-n  +  l) 


360  ALGEBRA. 

(5)  A  bag  contains  5  white  and  6  black  balls.     If  5  balls 

are  drawn  in  succession,  and  no  one  of  them  replaced, 

what  is  the  probability  that  the  first  three  will  be 

white,  and  the  fourth  and  fifth  black  ? 

The  separate  chances  for  the  5  simple  events  are  respectively 
XT'  T^i  I'  f '  f-  Hence,  the  chance  for  the  compound  event  is 
A  X  A  X  f  X  I  X  f  =  jh-  ^^s. 

(6)  Find  the  probability  in  the  last  example  that  the  5 

balls  dra.wn  will  be  3  white  and  2  black  balls. 

Here  the  chance  required  is  that  3  white  and  2  black  should 

be  drawn  not  in  any  assigned  order,  as  in  the  last  case,  but  in 

any  possible  order.     Now  5  things,  of  which  3  are  alike  and  the 

other  2  alike,  may  be  arranged  (|  411)  in 

15 
-=—  =  10  ways. 

[3  |_2  •^.  -.       . 


(7)  Find  the  respective  probabilities  in  Examples  (5)  and 

(6)  if  after  each  drawing  the  ball  is  replaced. 

In  Ex.  (5),    {^\fx{j\f.  A71S. 

In  Ex.  (6),    10  X  ( /t)='  X  (Af .  Ans. 

(8)  A  purse  contains  9  silver  dollars  and  1  gold  eagle,  and 

another  contains  10  silver  dollars.  If  9  coins  are 
taken  out  of  the  first  purse  and  put  into  the  second, 
and  then  9  coins  are  taken  out  of  the  second  and  put 
into  the  first  purse,  which  purse  now  is  the  more 
likely  to  contain  the  gold  coin  ? 

The  gold  eagle  will  not  be  in  the  second  purse  unless  it  (i.) 
was  among  the  9  coins  taken  out  of  the  first  and  put  into  the 
second  purse,  (ii.)  and  not  among  the  9  coins  taken  out  of  the 
second  and  put  into  the  first  purse.  The  chance  of  (i.)  is  ■^^,  and 
when  (i.)  has  happened  the  chance  of  (ii.)  is  |^.  Hence,  the 
chance  of  both  happening  is  y®^  X  xf  =  j^.  Therefore,  the  chance 
that  the  eagle  is  in  the  second  purse  is  -^g,  and  the  chance  that 
it  is  in  the  first  purse  is  1  —  -^^  =  -^f.  Since  ^^  is  greater  than 
j^^,  therefore  the  gold  coin  is  more  likely  to  be  in  the  first  purse. 


CHANCE.  361 


(9)    In  a  bag  are  2  red  and  3  white  balls.     A  is  to  draw  a 

ball,  then  B,  and  so  on  alternately ;  and  whichever 

draws  a  white  ball  first  is  to  receive  $  10.    Find  their 

expectations. 

A's  chance  of  drawing  a  white  ball  at  the  first  trial  is  |.  B's 
chance  of  having  a  trial  is  equal  to  A's  chance  of  drawing  a  red 
ball  =  f .  In  case  A  drew  a  red  ball  there  would  be  1  red  and  3 
white  balls  left  in  the  bag,  and  B's  chance  of  drawing  a  white 
ball  would  be  f .  Hence,  B's  chance  of  having  the  trial  and 
drawing  a  white  ball  is  f  X  f  =  x\ ;  and  B's  chance  of  drawing 
a  red  ball  is  f  x  }  =  yV- 

A's  chance  of  having  a  second  trial  is  equal  to  B's  chance  of 
drawing  a  red  ball  =  ^jj.  In  case  B  drew  a  red  ball  there  would 
be  3  white  balls  left,  and  A's  chance  of  drawing  a  white  ball 
would  be  certainty,  or  1, 

A's  chance,  therefore,  is  |  +  ^^  =  -j^ ;  and  B's  chance  is  y^^. 

A's  expectation,  then,  is  $7,  and  B's  $3. 

435.  In  general,  when  it  is  required  to  find  which  of  two 
doubtful  events  is  more  likely  to  happen,  it  is  necessary  to 
find  their  respective  chances,  and  then  to  compare  the 
results  obtained. 

(1)  In  one  throw  with  two  dice  which  sum  is  more  likely 

to  be  thrown,  9  or  12  ? 

Out  of  the  36  possible  ways  of  falling, /our  give  the  sum  9 
(namely,  6  +  3,  3  +  6,  5  +  4,  4  +  5),  and  onlg  one  way  gives  12 
(namely,  6  +  6).  Hence,  the  chance  of  throwing  9  is  four  times 
as  good  as  that  of  throwing  12. 

(2)  With  three  dice  what  are  the  relative  chances  of  throw- 

ing a  doublet  and  a  triplet  ? 
The  chance  of  throwing  a  doublet  is 

The  chance  of  throwing  a  triplet  is 
6  X  1  X  1_^   1 

Hence,  the  chance  of  a  doublet  is  15  times  that  of  a  triplet. 


3G2  ALGEBRA. 


(3)  A  bag  contains  1  black  and  4  white  balls,  and  another 

bag  contains  7  black  and  3  white  balls.     If  a  person  . 
draws  a  ball  from  one  of  the  bags,  (i.)what  is  the 
chance  that  it  be  a  white  ball  ?  (ii.)  what  is  the  ratio 
of  the  chance  of  its  being  drawn  from  the  first  bag  to 
that  of  its  being  drawn  from  the  second  bag  ? 
The  person  (so  far  as  we  know)  is  as  likely  to  choose  one  bag 
as  the  other.     Hence, -the  chance  of  his  choosing  the  first  bag  is 
^;  and  the  chance  of  his  drawing  a  white  ball  from  the  first  bag 
is  f.     Therefore,  the  chance  of  drawing  a  white  ball  from  the 
first  bag  is  J  X  f  =  |.     In  the  same  way,  the  chance  of  drawing 
a  white  ball  from  the  second  bag  is  found  to  be  J  X  jjj  =  2V 

Therefore,  the  chance  of  drawing  a  white  ball  is  f  +  2^^  =  |i ; 
and  the  ratio  of  the  separate  chances  is  8 :  3. 

(4)  Suppose  in  the  last  example  that  at  the  first  trial  a 

white  ball  is  actually  drawn.     What  are  now  the 

chances  that  it  came  from  the  first  bag,  and  from  the 

second,  respectively? 

Let  X  and  y  represent  the  chances  required. 

Then,  by  Ex.  (3),  ^  =  ?-. 

y    3 

Also,  X  +  y  =  1, 

since  the  ball  must  have  come  from  one  or  the  other  bag. 
The  solution  of  these  equations  gives 

x  =  j\,      and      3/  =  xV 

436.  From  Examples  (3)  and  (4)  it  will  be  seen  that, 
IX.  Jf  a  doubtful  event  Tnay  happen  in  some  one  of 
several  ways,  the  actual  happening  of  the  event  changes  its 
prohahilitg,  and  the  separate  probabilities  of  the  several  wags 
of  happening,  in  the  same  ratio ;  and  this  ratio  is  the  recip- 
rocal of  the  fraction  that  expresses  the  chance  of  the  event 
before  it  actually  happens. 

Thus,  in  Ex.  (3),  the  chance  of  the  event  before  it  happened,  and 
the  chance  of  the  two  separate  ways  of  happening,  were  found  to  be 


CHANCE.  363 


Ih  -iu^  ^^ff ;  ^^  ■'^^-  (^)  ^^  "^^^  shown  that  after  the  event  happened  these 
chances  became  1,  -^j,  j\,  respectively.  The  values  1,  j\,  ^j  are  ob- 
tained by  multiplying  |i  2%,  ij  ^7  tt!  ^^^^  ^^'  ^7  ^^^^  reciprocal  of^. 

Evidently,  the  happening  of  the  event  must  change  to  unity  the 
chance  of  the  event ;  and  must  therefore  increase  to  unity  the  sum  of 
the  separate  chances. 

So  long,  however,  as  the  only  additional  knowledge  about  the  event 
is  the  fact  that  it  has  happened,  the  relative  probabilities  of  the  sepa- 
rate ways  of  happening  remain  unchanged.  Therefore,  the  several 
fractions  which  before  expressed  the  probabilities  of  the  separate 
ways  of  happening  must  now  be  multiplied  by  the  same  factor,  and 
that  factor  is  the  reciprocal  of  the  fraction  that  expressed  the  proba- 
bility of  the  event  before  it  happened. 

Exercise  CXXL 

1.  The  chance  that  A  can  solve  a  certain  problem  is  J,  and 

the  chance  that  B  can  solve  it  is  f .     What  is  the 
chance  that  the  problem  will  be  solved  if  both  try  ? 

2.  What  is  the  chance  of  throwing  at  least  one  ace  in  2 

throws  with  one  die  ? 

3.  If  n  coins  are  tossed  up,  what  is  the  chance  that  one, 

and  only  one,  will  turn  up  head? 

4.  What  is  the  chance  of  throwing  double  sixes  at  least 

once  in  3  throws  with  2  dice  ? 

5.  A  copper  is  tossed  3  times.     Find  the  odds  that  it  will 

fall: 

(i.)  Head  and  two  tails  without  regard  to  order, 
(ii.)  Head,  tail,  head. 

6.  If  a  copper  is  tossed  4  times,  find  the  odds  that  it  will 

fall  2  heads  and  2  tails  sooner  than  4  heads. 

7.  If  from  a  lottery  of  30  tickets,  marked  1,  2,  3, ,  four 

tickets  are  drawn,  what  is  the  chance  that  1  and  2 
will  be  among  them  ? 


304  ALGEBRA. 


8.  If  2  coppers  are  tossed  3  times,  find  the  odds  that  they 

will  fall  2  heads  and  4  tails. 

9.  There  are  10  tickets,  five  of  which  are  numbered  1,  2, 

3,  4,  5,  and  the  other  five  are  blank.  Find  the 
chance  that  the  sum  of  the  numbers  on  the  tickets 
drawn  in  3  trials  will  be  10,  one  ticket  being  drawn 
and  then  replaced  at  each  trial  ? 

10.  Find  the  chance  in  Ex.  9  if  the  tickets  are  not  replaced. 

11.  A  bag  contains  4  white  and  6  red  balls.     A,  B,  and  C 

draw   each  a  ball,   in    order,   replacing.     Find   the 
chance  that  they  have  drawn  : 
(i.)  Each  a  white  ball, 
(ii.)  A  and  B  white,  C  red. 
(iii.)  Two  white  and  one  red. 

12.  Find  the  answer  to  Ex.  11  if  the  balls  are  not  replaced. 

13.  A  draws  4  times  from  a  bag  containing  2  white  and  8 

black  balls,  replacing.     Find  the  chance  that  he  will 
have  drawn : 
(i.)  Two  white,  two  black, 
(ii.)  Not  less  than  two  white, 
(iii.)  Not  more  than  two  white, 
(iv.)  One  white,  three  black. 

14.  Find  the  odds  against  throwing  one  of  the  two  numbers 

7  or  11  in  a  single  throw  with  2  dice. 

15.  If  a  copper  is  tossed  5  times,  what  is  the  chance  that  it 

will  fall  heads  either  2  times  or  else  3  times  ? 

16.  Find  the  same  chance  if  the  copper  is  tossed  6  times. 

17.  In  one  bag  are  10  balls  and  in  another  6 ;  and  in  each 

bag  the  balls  are  marked  1,  2,  3,  etc.  What  is  the 
chance  that -on  drawing  one  ball  from  each  bag  the 
two  balls  will  have  the  same  number  ? 


CHANCE.  365 


18.  A  bag  contains  n  balls.     A  person  takes  out  one  ball 

and  then  replaces  it.  He  does  this  n  times.  What 
is  the  chance  that  he  has  had  in  his  hand  every  ball 
in  the  bag  ? 

19.  If  on  an  average  9  ships  out  of  10  return  safe  to  port, 

what  is  the  chan&e  that  out  of  5  ships  expected  at 
least  3  will  return  ? 

20.  What  is  the  chance  of  throwing  double  sixes  at  least 

once  in  3  throws  with  a  pair  of  dice  ? 

21.  What  is  the  chance  of  throwing  15  in  one  throw  with 

3  dice  ? 

22.  In  5  throws  with  a  single  die  what  is  the  chance  of 

throwing  an  ace  : 
(i.)  Three  times  exactly. 
(ii.)  Not  less  than  three  times, 
(iii.)  Not  more  than  three  times. 

23.  In  a  bag  are  3  white,   5  red,  and  7  black  balls,  and 

a  person  draws  three   times,  replacing.      Find  the 
chance  that  he  will  have  drawn : 
(i.)  A  ball  of  each  color, 
(ii.)  Two  white,  one  red. 
(iii.)  Three  red. 
(iv.)  Two  red,  one  black. 

24.  A  and  B  play  at  chess,  and  A  wins  on  an  average  2 

games  out  of  3.  Find  the  chance  of  A's  winning  ex- 
actly 4  games  out  of  the  first  6,  drawn  games  being 
disregarded. 

25.  A  and  B  engage  in  a  game  in  which  A's  skill  is  to  B's 

as  2 :  3.  Find  the  chance  of  A's  winning  at  least  2 
games  out  of  the  first  5,  drawn  games  not  being 
counted. 


366  ALGEBKA. 


26.  The  skill  of  A  is  double  that  of  B.     Find  the  odds 

against  A's  winning  4  games  before  B  wins  2. 

27.  If  B's  skill  in  a  certain  game  is  equal  to  three-fifths  of 

A's,  find  A's  chance  of  winning  5  games  out  of  8. 

28.  A  bag  contains  4  red  balls  and  2  others,  each  of  which 

is  equally  likely  to  be  red  or  white.  Three  times  in 
succession  a  ball  is  drawn  and  replaced.  Find  the 
chance  that  all  the  draAvn  balls  are  red. 

29.  A  man  has  left  his  umbrella  in  one  of  3  shops  which  he 

visited  in  succession.  He  is  in  the  habit  of  leaving 
it,  on  an  average,  once  every  4  times  that  he  goes  to 
a  shop.  Find  the  chance  that  he  left  it  in  the  first, 
second,  and  third  shops,  respectively. 

30.  A  bets  B  ^  10  to  $  1  that  he  will  throw  heads  at  least 

once  in  3  trials.  What  is  B's  expectation?  What 
would  have  been  a  fair  bet  ? 

31.  A  draws  5  times  (replacing)  from  a  bag  containing  3 

white  and  7  black  balls ;  every  time  he  draws  a  white 
ball  he  is  to  receive  $1,  and  every  time  he  draws 
a  black  ball  he  is  to  pay  50  cents.  What  is  his 
expectation  ? 

32.  From  a  bag  containing  2  eagles,  3  dollars,  and  3  quarter- 

dollars,  A  is  to  draw  one  coin  and  then  B  three  coins; 
and  A,  B,  and  0  are  to  divide  equally  the  value  of 
the  remainder.     What  are  their  expectations  ? 

33.  A,  B,  and  C,  staking  each  $5,  draw  from  a  bag  in  which 

are  4  white  and  6  black  balls,  each  drawing  in  order, 
and  the  whole  sum  is  to  be  received  by  him  who  first 
drawg  a  white  ball.     What  are  their  expectations : 
(i.)  Replacing  the  balls, 
(ii.)  Not  replacing  the  balls. 


CHAPTER    XXIV. 
Formulas. 

Simple    Interest. 

437.  lim  Interest 

The  principal  be  represented  by  P, 
interest  on  §  1  for  one  year  by  r, 
amount  of  §  1  for  one  year  by  i?, 
number  of  years  by  n, 

£Lmount  of  P  for  n  years  by        A,    - 
Then  i2  =  l  +  r, 

Simple  interest  on  P  for  a  year     =  Pr, 
Amount  of  P  for  a  year  =  PR, 

Simple  interest  on  P  for  n  years  =  Pnr, 
Amount  of  P  for  n  years  =  P'(l  +  nr). 

That  is,  ^  =  P(l4-wr). 

438.  When  any  three  of  the  quantities  A,  P,  n,  r  are 
given,  the  fourth  may  be  found. 

Ex.  Required  the  rate  when  $500  in  4  years  at  simple 
interest  amounts  to  $610. 

r  is  required,  A,  P,  n  are  given. 

A  =  P{1+  nr), 
or  A  =  F+  Pnr. 

.'.  Pnr  =  A-P, 

.     ^^A-p^m-m_^^^ 

Pn  2000 

5|  per  cent.  Ans. 


368  ALGEBRA. 

439.  Since  P  will  in  n  years  amount  to  A,  it  is  evident 
that  P  at  the  present  time  may  be  considered  equivalent  in 
value  to  A  due  at  the  end  of  w  years ;  so  that  P  may  be 
regarded  as  i\iQ  present  worth  of  a  given  future  sum  A.  ^ 

Ex.  Find  the  present  worth  of  §600,  due  in  2  years,  the 
rate  of  interest  being  6  per  cent. 
A  =  P(1  +nr). 

•    P  -  ^ ^  ^^^  -  $ 535.71. 


1  +  wr      1  +  12 


Compound  Interest. 

440.   When  compound  interest  is  reckoned  payable  annu- 
ally, 

The  amount  of  P  dollars  in 

1  year  is        P(l  +  r)  =  Pi?, 

2  years  is  PR  (1  +  r)  =  PP^ 
n  years  =  PP**. 

That  is,  A^PR-. 

Hence,  also,  P=—~. 

When  compound  interest  is  payable  semi-annually, 
The  amount  of  P  dollars  in 

^year-^P^l  +  0 
lyear  =p(^l+|y, 
n  years  =  P  (  1  -f  - 

„,  .  .         _  /  .         r\2n 

That  IS,  \      ■  9 

When  the  interest  is  payable  quarterly, 

4 


FORMULAS.  369 


When  the  interest  is  payable  monthly, 


When  interest  is  payable  q  times  a  year 

( 


A=p[i^^-r 


1) 

Ex.  Find  the  present  worth  of  $500,  due  in  4  years,  at  5 
per  cent  compound  interest. 
A  =  P{l-{-rf. 

Sinking  Funds. 

441.    If  the  sum  set  apart  at  Ihe  end  of  each  year  to  be 
put  at  compound  interest  be  represented  by  8,  then, 
The  sum  at  the  end  of  the 
first  year      =  8, 
second  year  =  8-\-  8R, 
third  year    -=  8+ 8E-\- 8R'', 

nth  year       =  8  +  8R -^  81^  + +  8Br-^. 

That  is,  the  amount  A  =  8 -\- 8R -\- 8R^ -{- +xS'^-\ 

AR  =  8R-\-8R''^8R^+ +  8Rr. 

:.AR-A  =  8R^-8. 
^^_8{R--l) 
R-l 

A  =  ^(^-'l 
•  r 

(1)    If  $  10,000  be  set  apart  annually,  and  put  at  6  per  cent 
compound  interest   for  10  years,  what  will  be  the 

amount  ? 

S(R''-1)     1 10,000  (1.06»0-1) 
r         ~  .06 

By  logarithms  the  amount  is  found  to  be  $131,710  (nearly). 


370  ALGEBRA. 

(2)    A  county  owes  $60,000.     What  sum  must  be  set  apart 
annually,  as  a  sinking  fund,  to  cancel  the  debt  in  10 
years,  provided  money  is  worth  6  per  cent  ? 
^^  _A^  ^  $60,000  X. 06  ^  ^,,,,  ^^^^^^  ) 

Note.  The  amount  of  tax  required  yearly  is  $  3600  for  the  interest 
and  $4555  for  the  sinking  fund;  that  is,  $8155. 

Annuities. 

442.    A  sum  of  money  that  is  payable  yearly,  or  in  parts 
at  fixed  periods  in  the  year,  is  called  an  aimiiity. 

I.    To  find  the  amount  of  an  unpaid  annuity  when  the 
intei^estf  time,  and  rate  -per  cent  are  given. 
The  sum  due  at  the  end  of  the 
first  year      =  /§■, 
second  year  =  /S'+  SB,, 
third  year    =  8 -^  BR -\- BB? , 
nth  year       =  /S'-f  /^i?  +  BIv -^ +  SE^-^. 

That  is,  A  =  ^(-^~^).  §  441. 

Ex.  An  annuity  of  §1200  was  unpaid  for  6  yeai-s.     What 

was  the  amount  due  if  interest  be  a  eckoned  at  6  per 

cent? 

^^g(ig^)^$120O(L06«-l)^^33^^_ 

XL    To  find  the  present  worth  of  an  annuity  when  the  time 
it  is  to  continue  and  the  rate  per  cent  are  given. 

Let  P  denote  the  present  worth.     Then  the  amount  of  P  for  n 
years  will  be  equal  to  A,  the  amount  of  the  annuity  for  n  years. 

But  the  amount  of  P  for  n  years 

=  P(l+r)«=Pi>:«, 

and  ^^^(|^.  Hll. 

JXi  —  J.  . 


FORMULAS.  371 


This  equation  may  be  written 

B-l^     E- 

If  the  annuity  is  perpetual,  the  fraction 

E^-l 

E^ 

approaches  to  unity  as  its  limit. 

r.P     limit  of  J     X^-^- 

E-1         i2«        E 

8        S 
-1     r 

(1)    Find  the  present  worth  of  an  annual 

pension 

of  $105. 

for  5  years,  at  4  per  cent  interest. 

'          E-""  E-l 

^|10r,     ]L04^1^^ 

l.O-l'     1.04-1     ^        ^         ^^ 

(2)    Find  the  present  worth  of  a  perpetual  scholarship  that 
pays  $300  annually,  at  6  per  cent  interest. 

P  =  ^  =  i^  =  $5000. 
r        .06       ^ 

III.  To  find  the  present  worth  of  an  annuity  thai  begins 
in  a  given  number  of  years,  when  the  thne  it  is  to  coTitinue 
and  the  rate  per  cent  are  given. 

Let  p  denote  the  number  of  years  before  the  annuity  begins,  and 
q  the  number  of  years  the  annuity  is  to  continue. 

Then  the  present  worth  of  the  annuity  to  the  time  it  terminates  is 
S         Ep+9  -  1 
Ep+9        E~1    ' 
and  the  present  worth  of  the  annuity  to  the  time  it  begins  is 
^      i2p-l 
Ep      E-1 

Hence.        p.  f^  x  ^^^^^  V  f"  X  ^^^^Y 
\Ep+9        E-1    J      \Ep      E-lj 

S     ^E^-l 


Ep+1      E-l 


372  ALGEBRA. 


If  the  annuity  is  to  begin  at  the  end  of  p  years,  and  to  be  per- 
petual, the  formula 

becomes  P=— — -= — rr  X        ~ 


IiP{Ii-l)         R^ 

R^  —  1 
And  since  the  limit  of  — - —  is  unity, 
Ri  ^ 

P=  the  limit  of ^ X  ^^^—1  = -^ 

Rp{R-l)        R^        Rp{R-l) 

(1)  Find  the  present  Avorth  of  an  annuity  of  $5000,  to  be- 
gin in  6  years,  and  to  continue  12  years,  at  6  per  cent 
interest. 


Rp+g  B-l 
p5000  L06^ 
1.0618  .06 


15000^1.061^-1^^ 


(2)    Find  the  present  wortli  of  a  perpetual  annuity  of  $  1000, 
to  begin  in  3  years,  at  4  per  cent  interest  ? 

P= ^ =  _li0^2_  ==  $22,225. 

Rp{R~l)     1.04=*  X. 04     ^ 

IV.    To  find  the  annuity  when  the  present  worth,  the  time, 
and  the  rate  'per  cent  are  given. 

j,_S(R--l) 
R»{R-\) 


(1)    What  annuity  for  5  years  will  $4675  give  when  inter- 
est is  reckoned  at  4  per  cent? 

;S'=  Fr  X  --^  =  ?4675x-.04x  -^^  =  ?1050. 
R»~l  1.04"- 1 


FORMULAS.  373 


Life  Insurance. 

443.  In  order  that  a  certain  sum  may  be  secured,  to  be 
payable  at  the  death  of  a  person,  he  pays  yearly  a  fixed 
premium. 

If  P  denote  the  premium  to  be  paid  for  n  years  to  insure  an 
amount  A,  to  be  paid  immediately  after  the  last  premium,  then 

A^i-ff^.  2  441. 

.  p_^(ig-l)^     Ar 

If  A  is  to  be  paid  a  year  after  the  last  premium,  then 
p__A{R-\)_       Ar 

Note.  In  the  calculation  of  life  insurances  it  is  necessary  to 
employ  tables  which  show  for  any  age  the  probable  duration  of  life. 

Bonds. 

444.  If  P  denote  the  price  of  a  bond  that  has  n  years  to 
run,  and  bears  r  per  cent  interest,  >iS'  the  face  of  the  bond, 
and  q  the  current  rate  of  interest,  what  interest  on  his  in- 
vestment will  a  purchaser  of  such  a  bond  receive  ? 

Let  X  denote  the  rate  of  interest  on  the  investment. 

Then  P(l  -\-xY  is  the  value  of  the  purchase  money  at  the  end  of 
n  years. 

Sr  (1  +  j)"-!  +  /Sr  (1  +  5')»'-2  + j^  Sr  ■\-  S  IB  the  amount  of  the 

money  received  on  the  bond  if  the  interest  received  from  the  bond  is 
put  immediately  at  compound  interest  at  q  per  cent. 

9. 

f  o       .a»  r/1 
.-.  1  +  a 


_{8  ST\{X^qY-\\\k 
U  P<1  J 

_fSg  +  Sr(l  +  q)^-Sr\k 
V  ^2  J' 


374  ALGEBRA. 


(1)  What  interest  will  a  person  receive  on  his  investment 

if  he  buys  a  4  per  cent  bond,  at  114,  that  has  26 
years  to  run,  and  if  money  is  worth  3^  per  cent? 
^  /3.5  +  4(1.035)^-4yV. 

V        3.99        ; 

By  logarithms,  l+x  =  1.033. 

That  is,  the  purchaser  will  receive  3J  per  cent  for  his  money. 

(2)  At  what  price  must  7  per  cent  bonds  be  bought,,  run- 

ning 12  years,  with  the  interest  payable  semi-annu- 
ally, in  order  that  the  purchaser  may  receive  on  his 
investment  5  per  cent,  interest  eemi-annual,  which  is 
the  current  rate  of  interest  ? 

.  j,^Sq  +  Sr{l  +q)n-Sr 

In  this  case  8=  100  ;  and,  as  the  interest  is  semi-annual, 
q  =  .025,   r  =  .035,   n  =  24,   x==  .025. 

Hence.  p^  2.5  +  3.5(1.025^-3.5^ 

.025(1.025)*-^ 
By  logarithms,     P=  118. 

Exercise  CXXII. 

1.  In  how  many  years  will  $  100  amount  to  $  1050,  at  5 

per  cent  compound  interest  ? 

2.  In  how  many  years  will  ^A  amount  to  $B  (i.)  at  sim- 

ple interest,  (ii.)  at  compound  interest,  r  and  i?  being 
used  in  their  usual  sense  ? 

3.  Find  the  difference  (to  "five  places  of  decimals)  between 

the  amount  of  $  1  in  2  years,  at  6  per  cent  compound 
interest,  according  as  the  interest  is  due  yearly  or 
monthly. 

4.  At  5  per  cent,  find  the  amount  of  an  annuity  A  which 

has  been  left  unpaid  for  4  years. 


FORMULAS.  375 


5.  Find  the  present  value  of  an  annuity  of  $100  for  5 

years,  reckoning  interest  at  4  per  cent  ? 

6.  A  perpetual  annuity  of  $1000  is  to  be  purchased,  to 

begin  at  the  end  of  10  years.  If  interest  is  reckoned 
at  3 J  per  cent,  what  should  be  paid  for  it? 

7.  A  debt  of  $1850  is  discharged  by  two  payments  of 

$  1000  each,  at  the  end  of  one  and  two  years.  Find 
the  rate  of  interest  paid. 

8.  Reckoning  interest  at  4  per  cent,  what  annual  premium 

should  be  paid  for  30  years,  in  order  to  secure  $2000 
to  be  paid  at  the  end  of  that  time,  the  premium  being 
due  at  the  beginning  of  each  year? 

9.  An  annual  premium  of  $  150  is  paid  to  a  life-insurance 

company  for  insuring  $5000.  If  money  is  worth  4 
per  cent,  for  how  many  years  must  the  premium  be 
paid  in  order  that  the  company  may  sustain  no  loss? 

10.  What  may  be  paid  for  bonds  due  in  10  years,  and  bear- 

ing semi-annual  coupons  of  4  per  cent  each,  in  order 
to  realize  3  per  cent  semi-annually,  if  money  is  worth 
3  per  cent  semi-annually  ? 

11.  When  money  is  worth  2  per  cent  semi-annually,  if  bonds 

having  12  years  to  run  and  bearing  semi-annual  cou- 
pons of  3 J  per  cent  each,  ar£  bought  at  114 J,  what 
per  cent  is  realized  on  the  investment  ? 

12.  If  $  126  is  paid  for  bonds  due  in  12  years,  and  yielding 

3^  per  cent  semi-annually,  what  per  cent  is  realized 
on  the  investment,  provided  money  is  worth  2  per 
cent  semi-annually? 


376  ALGEBRA. 


Binomial  Formula. 
445.    By  performing  the  indicated  multiplication, 

(x -\- a)  (x -}- b)  =^  x^  -\- {a -\- b)  X -{-  ab. 
Multiply  each  side  by  a;  -f  ^,  and  the  result  is 

{x-\-a)  {x-\-b)  {x-\-c) = a^+  (a+b)  x^     +  abx 

+  cx^     -\-        (ac-^bc)  x-{-abc 

= af^~{-  (a-{-b-\-c)  y?-\-  {ab^ac-\-bc)  x-^abc 
From  this  result  certain  law8  are  to  be  observed : 

I.  The  ^number  of  terms  is  one  more  than  the  number  of  factors 
on  the  left  side. 

II.  The  exponent  of  x  in  the  first  term  is  the  same  as  the  number 
of  factors,  and  decreases  by  1  in  each  succeeding  term. 

III.  The  coefficient  of  x  in  the  first  term  is  unity  ; 
in  the  second  term,  the  sum  of  a,h,  c\ 

in  the  third  term,  the  sum  of  the  products  two  and  two,  ah,  ac,  he ; 
and  the  fourth  term  is  the  product  ahc. 

Do  the  same  laws  hold,  whatever  be  the  number  of  factors  ? 
Suppose  that  these  laws  hold  for  r  factors,  so  that 

{x  +  a){x  +  h) {x  -\-m)  =  c(f  -[■p^x^-'^  +^2^*""^  +  p^x!^'^  + -\-  pr 

where  p^  stands  for  a  +  &  + +  m,  the  sum  of  the  second  terms ; 

Pj  stands  for  ab  +  ac  + ,  the  sum  of  the  products,  two  and 

two ; 

Pg  stands  for  ahc  +  ahd  + ,  the  sum  of  the  products,  three  and 

three, 
p^  stands  for  abed ,  the  product  of  the  r  letters. 

Multiply  by  another  factor  {x  +  n),  and  the  product  of  the  r  +  1 
factors  is 

af+i+    jj^ar     +      p^vi^-^       +      Pga;'-^       + +     p^x 

+     nxf       +      p^na^-i     +      j»,waf—2    + +     p^-xnx      -^-prn 

=a;»'+i+(Pi+n)a^+(p24-p^n)a;'-i+(;)3+P2^)af-2+ +{pr+pr-\n)x->rprn 

Here  the  laws  I.  and  II.  evidently  hold ;  and  as  to  the  coefficients 
p,  +n=a  +  6+ +  m  +  n,  the  sum  of  the  r  +  1  letters  ; 


FORMULAS.  377 


Pi  +  Pi'>^  =  (a6  +  ac  + )  +  {an  +  bn  + +  mn),  the  sum  of  the 

products,  two  and  two  ; 

Pa  +1^2^  ^  (^^<^  +  ^^^  + )  +  ('^^'^  +  ^^^  + )>  ^^®  ^^^  ^^  ^^^ 

products,  three  and  three. 

p^n  =  a6c mn,  the  product  of  the  r  +  1  letters.     Hence, 

The  laws  hold  for  r  +  1  factors  if  they  hold  for  r  factors. 

But  they  have  been  shown  to  hold  for  three  factors,  therefore  they 

hold  for  four  factors,  and  therefore  for  Jive  factors ;  and  so  on,  for  any 

number  of  factors.* 

446.   When  the  factors  in   the   preceding  proof  are  all 

equal,  so  that  b,  c,  d, n,  are  each  equal  to  a,  the  left 

side  of  the  equation  becomes 

(a;  +  a){x  +  a) taken  n  times ;  that  is,  (x  +  ay*. 

On  the  right  side, 

jPi  =  a  +  a  + =  a  taken  n  times  =  n/i ; 

p^  =  aa  +  aa  + =  a"  taken  as  many  times  as  there  is  a  choice 

of  2  letters  from  n  letters, 

that  is,  P  =  n{n-l)    ,  „  ^^^ 

'        1X2        '  ^ 

^3  =»  aaa  +  aaa  + =  a^  taken  as  many  times  as  there  is  a  choice 

of  3  letters  from  n  letters, 

thati«,  P.  =  Ei^^nfe^„3.  J  415. 


Prn  =  ax  ax  a =  a  taken  n  times  as  a  factor  =  a". 

.-.  {x  +  a>  =  ai>*  +  naa^-''-  +  ^  (^  ~  •*•)  a*x«-» 
1x2 

^n(n-l)(n-2)^3^.3_^ ^  ^„ 

1x2x3 

447.  The  expression  on  the  right  side  is  called  the  expan- 
sion of  (x  +  a)**. 

If  a  and  x  be  interchanged,  the  expansion  will  proceed 
by  ascending  powers  of  x,  as  follows  : 

{a  +  a;)«  =  a"  +  ?ia«-ia;  +  'n{n  —  l)  ^n-2^  ^ ^  nax''-'^  +  .r«. 

1  X  ^ 

*  A  proof  of  this  kind  is  called  mathematical  induction. 


378  ALGEBRA. 


If  a  =  1,  then, 
{l+x)^=l+nx  +  ^  ^^^  ~^^  x^  + +  ?ia;«-i  +  a;«. 

If  X  be  negative,  the  odd  powers  of  x  will  be  negative  and  the  even 
powers  positive. 

(a  -  xY  =  a"  -  na"-!  x  +  — ^^i:=iD  a«-2  a;2 
1x2 

_n(n-l)(n-2)^„_3^3  + 

1x2x3 


448.  It  will  be  observed  that  the  last  factor  in  the  de- 
nominator of  the  coefficient  is  1  less  than  the  number  of  the 
term,  and  is  the  same  as  the  exponent  of  the  second  letter ; 
also,  that  the  last  factor  of  the  numerator  of  the  coefficient 
is  found  by  subtracting  the  last  factor  in  the  denominator 
from  ^4-  1,  and  that  the  exponent  of  the  first  letter  is  found 
by  subtracting  the  exponent  of  the  second  letter  from  n. 
So  that, 

The  rtli  (or  general)  term  in  the  expansion  of  (a  +-a;)"  is 

n{n-V) (n-r  +  2)  ^„.^+i^^,i 

1X2 (r-1) 

Thus,  the  third  term  of  (a  +  xf^  is 

20  X  19  ^18^^  190^18 a^^ 
1X2 

449.  The  -coefficient  of  the  rth  term  from  the  beginning 
is  equal  to  the  coefficient  of  the  rth  term  from  the  end. 

For,  the  coefficient  of  the  rth  term  from  the  beginning  is 

n(n— 1) (n-r  +  2)      ^^    n(n-l) (n  — r  +  2). 

1X2 r-1  '       ^  £^3  ' 

and  this  becomes,  when  both  terms  are  multiplied  by  \n  —  r  +  1, 

|r  —  1  |n  —  r  +  1 


FORMULAS.  379 


The  coefficient  of  the  rth  term  from  the  end,  which  is  the  {n—r-\-2)th 
term  from  the  beginning,  is 

n(n  — 1) r 

\n  —  r+  1 

and  this  also  becomes,  when  both  terms  are  multiplied  by  |r  — 1, 

Ir  —  1   Iti  — r  +  1 


450.  It  will  be  evident  from  §  417,  that  in  the  expansion 
of  (a  +  xy,  the  middle  term  will  have  the  greatest  coeffi- 
cient when  n  is  even ;  and  when  n  is  odd  the  two  middle 
terms  will  have  equal  coefficients,  and  these  will  be  the 
greatest. 

Exercise  CXXIII. 
Expand : 

1.  (l4-2a;/,  3.    {2x-2>y)\  5.    (l-^Y 

2.  (a;-3)«.  4.    (2-x)\  6.    (l-^Y 

7.  Find  the  fourth  term  of  (2  a:  —  5?/)^^. 

8.  Find  the  seventh  term  of  f--f-^)  . 

9.  Find  the  twelfth  term  of  {a?  -  ax)^\ 

10.  Find  the  eighth  term  of  {bx^y  -  2xf)\ 

11.  Find  the  middle  term  of  (-J^'ts. 

12.  Find  the  middle  term  of  /"--^Y*^. 

13.  Find  the  two  middle  terms  of  (-  -  ^Y. 


380  ALGEBRA. 


14.  Find  tlie  rth  term  of  (2  a  +  a:)**. 

15.  Find  the  rth  term  from  the  end  of  (2  a  -f  rr)". 

16.  Find  the  (r  +  ^)th  term  of  {a  +  x^. 

17.  Find  the  middle  term  of  {a-^xf'', 

18.  Expand  (2a  +  :i-)^,  and  find  the  sum  of  the  terms  if 


Fres^uoork  by  EockweU  <&  Churchill. 


GINN,   HEATH,   &'   CO:S  PUBLICATIONS. 


MATHEMATICS. 


Wentujorth's  Elements  of  Plane  and  Solid  Ge- 

OMETRY.  By  George  A.  Wentvvoktii,  Phillips  Academy,  Exeter, 
i2mo.  Half  morocco.  400  pages.  Mailing  price,  $1.45;  Introduc- 
tion, J^i.oo;   Exchange,  60  cts. 


Wentmorth's  Elements  of  Plane  Geometry. 

l2mo.      250  pages.      Mailing   Price,   85   cts.;    Introduction,   75   cts.; 
Exchange,  40  cts. 

This  work  is  based  upon  the  assumption  that  Geometry  is  a 
branch  of  practical  logic,  the  object  of  which  is  to  detect,  and  state 
clearly  and  precisely,  the  successive  steps  from  premise  to  conclu- 
sion. 

In  each  proposition,  a  concise  statement  of  what  is  given  is 
printed  in  one  kind  of  type,  of  what  is  required  in  another,  and  the 
demonstration  in  still  another.  The  reason  for  each  step  is  indi- 
cated in  small  type,  between  that  step  and  the  one  following,  thus 
preventing  the  necessity  of  interrupting  the  process  of  demonstra- 
tion by  referring  to  a  previous  proposition.  The  number  of  the 
section,  however,  on  which  the  reason  depends,  is  placed  at  the 
side  of  the  page ;  and  the  pupil  should  be  prepared,  when  called 
upon,  to  give  the  proof  of  each  reason. 

A  limited  use  has  been  made  of  symbols,  wherein  symbols  stand 
for  words,  and  not  for  operations. 

Great  pains  have  been  taken  to  make  the  page  attractive.  The 
figures  are  large  and  elegant,  and  the  propositions  have  been  so 
arranged  that  in  no  case  is  it  necessary  to  turn  the  page  in  reading 
a  demonstration. 

A  large  experience  in  the  class-room  convinces  the  author  that, 
if  the  teacher  will  rigidly  insist  upon  the  logical  fonn  adopted  in 
this  work,  the  pupil  will  avoid  the  discouraging  difficulties  which 


A/A  THEM  A  TICS. 


usually  beset  the  beginner  in  geometry  ;  that  he  will  rapidly  develop 
his  reasoning  faculty,  acquire  facility  in  simple  and  accurate  expres- 
sion, and  lay  a  foundation  of  geometrical  knowledge  which  will  be 
the  more  solid  and  enduring  from  the  fact  that  it  will  not  rest  upon 
an  effort  of  the  memory  simply. 

Strong  evidence  of  the  merit  of  this  book  is  found  in  the  fact 
that  since  the  beginning  of  the  school  year,  1877-78,  it  has  been  intro- 
duced into  Thirty-six  Colleges  and  nearly  Four  Hundred  Preparatory 
Schools. 

Teachers  should  not  fail  to  examine  this  book  before  forming 
new  classes. 

TESTIMONIAI.S. 


Joseph  Ficklln,  Prof.  0/  Math., 
Univ.  of  Missouri  :  I  have  examined, 
with  considerable  care,  Wentworth's 
Geometry,  and  the  result  is  a  decidedly 
favorable  opinion  of  the  book.  Profes- 
sor Wentworth  is  evidently  a  practical 
teacher.  He  has  shown  in  the  execu- 
tion of  his  work  that  he  knows  just 
where  beginners  in  Geometry  encoun- 
ter difficulties,  and,  in  my  judgment, 
he  has  been  eminently  successful  in  his 
attempt  to  make  those  difficulties  dis- 
appear. 

Samuel  Hart,  Prof,  of  Math., 
Trinity  College  :  There  are  some 
things  in  Wentworth's  Geometry  which 
makes  it  specially  well  adapted  for 
use  in  Schools  and  Academies. 
The  clear  method  in  which  each 
proposition  is  presented  as  a  whole  ; 
the  manner  of  reproducing  the  state- 
ments of  former  theorems  to  which 
reference  is  made,  so  that  the  student 
is  almost  forced  to  recall  them;  the 
careful  and  precise  use  of  the  method 
of  limits  —  these  are  among  the  things 
which  will  tend  to  make  the  work  es- 
pecially serviceable.  And  I  am  satis- 
fied that  it  has  a  sufficiently  extensive 
view  of  Geometry  for  most  students, 


treated  in  a  way  which  will  help  both 
the  teaclier  and  the  taught. 
{^Nov.  22,  iSyg.) 

Selden  J.  Coffin,  Prof  of  Math., 
Lafayette  Coll.,  Pa. :  We  are  pleased 
with  the  simplicity  and  clearness  of  the 
demonstrations  in  Wentworth's  Geom- 
etry, and  have  adopted  it  as  our  text- 
book in  that  subject. 

E.  Otis  Kendall,  Prof  of  Math., 
Univ.  of  Pennsylvania  :  I  have  no  hes- 
itation in  saying  that  it  is  the  best  book 
for  beginners  that  I  have  ever  seen. 
The  demonstrations  are  rigorous,  the 
language  clear,  and  I  have  not  discov- 
ered any  defect  in  the  reasoning. 

W.  C.  Esty,  Prof  of  Math.,  Am- 
herst Coll.  :  It  is  a  step  in  advance  of 
the  text-books  of  its  kind.  It  must 
make  the  course  in  Geometry  easier  for 
both  teacher  and  pupil. 

John  R.  French,  Prof,  of  Math., 

Syracuse  Univ. :  The  distinctness  of 
its  statements,  and  the  clearness  and 
compactness  of  its  demonstrations,  are 
admirable.  We  purpose  to  test  it  in 
recitation-room  with  our  next  class. 


GINN,  HEATH,   &=    CO.'S  PUBLICATIONS. 


A.  S.  Hardy,  Prof,  of  Math.,  Dart- 
mouth Coll. :  I  regard  it  as  an  espe- 
cially attractive  book  to  students,  and 
welcome  it  among  books  of  its  class  as 
thoroughly  honest  and  founded  on  the 
only  method  suited  to  the  successful 
teaching  of  Elementary  Geometry. 

C.  A,  Waldo,  Instructor  in  Math., 
Wesley  an  Univ.:  It  stands  the  test  of 

use  in  class,  and  gives  the  best  of  re- 
sults. 

H.  N.  Wheeler,  Instructor  in 
Math.,  Harvard  Coll.  :  A  successful 
attempt  to  make  the  thorough  study 
of  Geometry  interesting  and  easy  for 
the  student,  and  the  teaching  of  the 
subject  a  pleasant  task  for  the  teacher. 

D.  McG.  Means,  MiddleburyColL, 
Vt. :  Wentworth's  Elements  of  Geom- 
etry seems  to  me  to  be  constructed  in 
accordance  with  recognized  psycho- 
logical laws.  The  attention  of  the  stu- 
dent is  economized  for  concentration 
upon  the  reasoning ;  and,  in  fact,  most 
difficulties  except  those  belonging  to 
the  subject  seem  to  be  removed. 
{^April  14, 1880.) 

McKendree  Petty,  Prof,  of 
Math.,  Univ.  of  Vermont:  I  find  the 
work  one  of  high  excellence  in  both 
matter  and  method.  I  am  glad  to  en- 
dorse its  merit,  and  feel  sure  that  its 
introduction  into  schools  will  contribute 
to  the  successful  study  of  this  branch 
of  learning. 

M.  W.  Humphreys,  Vanderbilt 
Univ. :  I  like  it  better  than  any  other 
that  I  am  acquainted  with,  intended 
for  schools  and  academies.  Our  Pro- 
fessor of  Mathematics  regards  it  as 
the  best  that  could  be  used  by  students 
preparing  to  enter  Vanderbilt  Univer- 
sity. 

William  V.  Brown,  Prof  of 
Math.,  Vanderbilt  Univ.,  Tenn. :  1  re- 
gard it  a  most  excellent  text-book,  and 


have  highly  commended  it  to  normal 
schools.     {Jan.  8, 1S80.) 

A.  L.  Nelson,  Prof  of  Math., 
Wash,  and  Lee  Univ.  :  I  have  used 
Wentworth's  Plane  and  Solid  Geome- 
try during  the  present  session,  and  I 
am  very  much  pleased  with  it.  It  is 
the  best  book  on  Geometry  I  ever  saw, 
in  the  style  of  its  demonstrations. 
{April  30, 1880.) 

G.  B.  McElroy,  Prof  of  Math., 
Adrian  Coll,,  Mich. :  It  is  a  very  de- 
cided improvement  on  all  other  text- 
books on  that  science.  I  intend  to  use 
it  in  my  classes  hereafter. 

E.  T.  Fristoe,  Prof,  of  Math., 
Columbian  Coll.,  Washington,  D.  C. : 
It  is  certainly  a  great  improvement 
on  the  old  wordy  method.  I  have 
placed  it  in  the  catalogue  for  introduc- 
tion at  the  next  session. 

S.  J.  Cunningham,  Prof  of 
Math.,  Swarthmore  Coll. :  I  am  glad 
to  express  my  hearty  appreciation  of  it. 
Students  cannot  fail  to  comprehend  it 
easily ;  and  I  am  sure  where  good  work 
can  be  done  with  any  Geometry,  Went- 
worth's cannot  fail  to  give  satisfaction. 

A.  P.  Peabody,  Harvard  Coll. : 
My  opinion  is  that  no  man  ever  had 
better  reasons  for  publishing  a  text- 
book. Taking  a  learner's  point  of  view, 
I  regard  a  knowledge  of  elementary 
Geometry  as  attainable  through  this 
book  at  much  less  cost  of  time,  and 
with  greater  precision  and  thorough- 
ness, than  by  means  of  any  other 
treatise  I  have  ever  seen.  The  demon- 
strations are  given  with  perfect  clear- 
ness. 

Thos.  Hill,  Ex-Pres.  of  Harvard 
Coll. :  It  is  the  best  Geometry  for 
Schools  I  ever  saw. 

C.  P.  P.  Bancroft,  Phillips  Acad., 
Andover,  Mass. :  As  1  examined  Went- 


MA  THEM  A  TICS, 


worth's  Geometry,  it  seemed  to  me  very 
clear  and  compact,  and  the  expedients 
for  expressing  graphically  the  terms 
and  the  relations  involved  in  the  sci- 
ence seemed  simple  and  sensible.  Orig- 
inal Vk'ork  is  also  well  provided  for. 
But  the  best  pledge  of  the  value^of  the 
book  is  its  author's  exceptional  success 
in  teaching  the  subject.  {April  i6, 
1880.) 

J.  B.  Sewall,  Prin.  of  Thayer 
Acad.,  Draintree,  Mass. :  We  have  used 
Wentworth's  Geometry  from  the  time 
of  its  publication,  and  are  so  com- 
pletely satisfied  with  it,  that  it  does  not 
now  seem  that  any  other  book  is  ever 
likely  to  take  its  place. 

H.  S.  Pershing,  Prof,  of  Math., 
Wilkesbarre  Acad.,  Pa.:  A  model  of 
clearness  and  conciseness  We  intend 
to  use  it  in  preparing  our  classes  for 
college. 

James  E.  Vose,  Prin.  of  Gushing 
Acad.,  Ashburnham,  Mass. :  I  have 
used  it  in  a  large  class,  the  past  year, 
and  it  is  the  best  Geometry  I  have  seen 
in  twenty  years'  experience. 

Sylvester  Dixon,  Prof  of  Math., 
N.  H.  Conference  Sent. :  I  have  been 
through  Wentworth's  Geometry  twice 
with  my  classes,  and  like  it  better  and 
better.  It  is  an  excellent  text-book,  and 
cannot  fail  to  be  extensively  used  when 
its  merits  become  known  to  teachers. 
{March  13, 1880.) 

J.  C.  Bartlett,  Prin.  Bristol  Acad., 
Taunton,  Mass. :  My  pupils  have  used 
the  book  with  gratifying  success.  Its 
many  original  advantages  recommend 
it  as  a  standard  text-book.  The  exam- 
ples are  excellent  for  developing  origi- 
nality in  the  pupils,  and,  if  I  mistake 
not,  in  the  teachers  also. 
{J^ch  1, 1880.) 

CharleaJSJ.  Jordan,  Prin.  Bangor 
High  Sch. :  It  was  last  year  introduced 


into  the  High  School,  and  gives  excel- 
lent satisfaction.  I  consider  it  the  best 
Geometry  for  high  schools  with  which 
I  am  acquainted.     {April  14, 1880.) 

Mary  H.  Ladd,  Prof  of  Math., 
Oshkosh  Nor7nal  Sch.,  Wis. :  What  I 
especially  admire  in  the  book  is  its 
teaching  quality  ;  it  seems  more  a  live 
teacher  than  a  text-book.  I  consider 
it  admirably  adapted  to  the  securing  of 
those  practical  results  and  that  disci- 
pline which,  with  most  books  in  the 
liands  of  the  average  student,  come 
only  from  the  efforts  of  the  teacher  in 
charge.     {March  j,  1880.) 

The  New  England  Journal  of 
Education :  We  do  not  need  to  re- 
fer to  the  title-page  of  this  book  to 
know  that  the  author  is  a  teacher.  In 
the  preface,  we  find  that  he  has  definite 
notions  of  what  a  Geometry  should  be, 
and  in  the  text  he  has  not  failed  to  carry 
out  his  ideas.  The  proof  of  proposi- 
tions by  direct  reference  to  the  princi- 
ples that  underlie  them,  instead  of  the 
circumlocution  too  often  met  with,  is 
an  excellent  feature  of  the  work.  The 
accuracy  of  the  definitions  is  also  wor- 
thy of  mention. 

Dascom  Greene,  Prof,  of  Math., 
Rensselaer  Polyt.  Inst.,  Troy  :  The  plan 
of  Wentworth's  Geometry  appears  to 
me  to  be  a  decided  improvement  in 
the  method  of  teaching  that  subject. 
{Sept.  ig,  iSyg.) 

D.  B.  Hagar,  Salem  Normal  Sch.: 
The  least  that  I  can  justly  say  in  regard 
to  Wentworth's  Geometry  is,  that  I 
know  of  no  work  superior  to  it  in  clear- 
ness, conciseness,  and  general  fitness 
for  use  in  schools.     {April  12, 1880.) 

Jas.  P.  Weston,  Westbrook  Sem., 
Me. :  We  think  it  a  very  superior  text- 
book, and  we  are  so  well  satisfied  with 
it,  that  we  shall  not  be  likely  to  exchange 
it  for  any  other.     {April  14, 18S0.) 


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